This is the easy version of the problem. The difference between the versions is in the number of possible operations that can be made. You can make hacks if and only if you solved both versions of the problem.

You are given a binary table of size n × m. This table consists of symbols 0 and 1.

You can make such operation: select 3 different cells that belong to one 2 × 2 square and change the symbols in these cells (change 0 to 1 and 1 to 0).

Your task is to make all symbols in the table equal to 0. You are allowed to make at most 3nm operations. You don't need to minimize the number of operations.

It can be proved that it is always possible.

Input

The first line contains a single integer t (1 ≤ t ≤ 5000) — the number of test cases. The next lines contain descriptions of test cases.

The first line of the description of each test case contains two integers n, m (2 ≤ n, m ≤ 100).

Each of the next n lines contains a binary string of length m, describing the symbols of the next row of the table.

It is guaranteed that the sum of nm for all test cases does not exceed 20000.

Output

For each test case print the integer k (0 ≤ k ≤ 3nm) — the number of operations.

In the each of the next k lines print 6 integers x_1, y_1, x_2, y_2, x_3, y_3 (1 ≤ x_1, x_2, x_3 ≤ n, 1 ≤ y_1, y_2, y_3 ≤ m) describing the next operation. This operation will be made with three cells (x_1, y_1), (x_2, y_2), (x_3, y_3). These three cells should be different. These three cells should belong into some 2 × 2 square.

Example

Input


5
2 2
10
11
3 3
011
101
110
4 4
1111
0110
0110
1111
5 5
01011
11001
00010
11011
10000
2 3
011
101


Output


1
1 1 2 1 2 2
2 
2 1 3 1 3 2
1 2 1 3 2 3
4
1 1 1 2 2 2 
1 3 1 4 2 3
3 2 4 1 4 2
3 3 4 3 4 4
4
1 2 2 1 2 2 
1 4 1 5 2 5 
4 1 4 2 5 1
4 4 4 5 3 4
2
1 3 2 2 2 3
1 2 2 1 2 2

Note

In the first test case, it is possible to make only one operation with cells (1, 1), (2, 1), (2, 2). After that, all symbols will be equal to 0.

In the second test case:

  * operation with cells (2, 1), (3, 1), (3, 2). After it the table will be: 
    
          
    011  
    001  
    000  
    

  * operation with cells (1, 2), (1, 3), (2, 3). After it the table will be: 
    
          
    000  
    000  
    000  
    




In the fifth test case:

  * operation with cells (1, 3), (2, 2), (2, 3). After it the table will be: 
    
          
    010  
    110  
    

  * operation with cells (1, 2), (2, 1), (2, 2). After it the table will be: 
    
          
    000  
    000  
    
import javax.naming.InitialContext;
import java.io.*;
import java.util.*;
import java.lang.*;
import java.math.*;
import java.text.DecimalFormat;
import java.lang.reflect.Array;
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigDecimal;
import java.util.concurrent.Phaser;

public class A {
    public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
    static int MOD = (int) (1e9 + 7);
    //static int MOD = 998244353;
    static long MOD2 = MOD * MOD;
    static FastReader sc = new FastReader();
    static int pInf = Integer.MAX_VALUE;
    static int nInf = Integer.MIN_VALUE;
    public static void main(String[] args) {
        int test = 1;
        test = sc.nextInt();
        while (test-- > 0) solve();
        out.flush();
        out.close();
    }
    static int[] di = new int[]{0,0,1,1};
    static int[] dj = new int[]{0,1,0,1};
    public static void solve() {
        int n = sc.nextInt();
        int m = sc.nextInt();
        int[][] a = new int[n][m];
        int count = 0;
        for(int i = 0; i < n; i++){
            char[] c = sc.next().toCharArray();
            for(int j = 0; j< m; j++){
                a[i][j] = c[j]=='1'?1:0;
            }
        }
        StringBuffer sb = new StringBuffer();
        for(int i = 0; i < n-1; i++){
            for(int j = 0; j < m-1; j++){
                int[][] p = new int[2][2];
                int c = 0;
                for(int k = 0; k < 4; k++){
                    int ni = i+di[k];
                    int nj = j+dj[k];
                    if(a[ni][nj]==0){
                        c++;
                    }
                }
                if(c==0){
                    count += 4;
                    sb.append((i+1)+" "+(j+2)+" "+(i+2)+" "+(j+2)+" "+(i+2)+" "+(j+1)+"\n");
                    sb.append((i+1)+" "+(j+1)+" "+(i+2)+" "+(j+2)+" "+(i+2)+" "+(j+1)+"\n");
                    sb.append((i+1)+" "+(j+2)+" "+(i+1)+" "+(j+1)+" "+(i+2)+" "+(j+1)+"\n");
                    sb.append((i+1)+" "+(j+2)+" "+(i+2)+" "+(j+2)+" "+(i+1)+" "+(j+1)+"\n");
                    for(int k = 0; k < 4; k++){
                        int ni = i+di[k];
                        int nj = j+dj[k];
                        a[ni][nj] = 0;
                    }
                }
                else if(c==1){
                    count += 1;
                    for(int k = 0; k < 4; k++){
                        int ni = i+di[k];
                        int nj = j+dj[k];
                        if(a[ni][nj]==1){
                            sb.append((ni+1)+" "+(nj+1)+" ");
                        }
                    }
                    sb.append("\n");
                    for(int k = 0; k < 4; k++){
                        int ni = i+di[k];
                        int nj = j+dj[k];
                        a[ni][nj] = 0;
                    }
                }
                else if(c==2){
                    count += 2;
                    int xi = 0;
                    int xj = 0;
                    for(int k = 0; k < 4; k++){
                        int ni = i+di[k];
                        int nj = j+dj[k];
                        if(a[ni][nj]==1){
                            xi = ni;
                            xj = nj;
                        }
                    }
                    for(int k = 0; k < 4; k++){
                        int ni = i+di[k];
                        int nj = j+dj[k];
                        if(ni==xi&&nj==xj){
                            continue;
                        }
                        if(a[ni][nj]==1){
                            a[ni][nj] = 0;
                        }
                        else{
                            a[ni][nj] = 1;
                        }
                        sb.append((ni+1)+" "+(nj+1)+" ");
                    }
                    sb.append("\n");
                    for(int k = 0; k < 4; k++){
                        int ni = i+di[k];
                        int nj = j+dj[k];
                        if(a[ni][nj]==1){
                            sb.append((ni+1)+" "+(nj+1)+" ");
                        }
                    }
                    sb.append("\n");
                    for(int k = 0; k < 4; k++){
                        int ni = i+di[k];
                        int nj = j+dj[k];
                        a[ni][nj] = 0;
                    }
                }
                else if(c==3){
                    count += 3;
                    int xi = 0;
                    int xj = 0;
                    for(int k = 0; k < 4; k++){
                        int ni = i+di[k];
                        int nj = j+dj[k];
                        if(a[ni][nj]==0){
                            xi = ni;
                            xj = nj;
                        }
                    }
                    for(int k = 0; k < 4; k++){
                        int ni = i+di[k];
                        int nj = j+dj[k];
                        if(ni==xi&&nj==xj){
                            continue;
                        }
                        if(a[ni][nj]==1){
                            a[ni][nj] = 0;
                        }
                        else{
                            a[ni][nj] = 1;
                        }
                        sb.append((ni+1)+" "+(nj+1)+" ");
                    }
                    sb.append("\n");

                    xi = 0;
                    xj = 0;
                    for(int k = 0; k < 4; k++){
                        int ni = i+di[k];
                        int nj = j+dj[k];
                        if(a[ni][nj]==1){
                            xi = ni;
                            xj = nj;
                        }
                    }
                    for(int k = 0; k < 4; k++){
                        int ni = i+di[k];
                        int nj = j+dj[k];
                        if(ni==xi&&nj==xj){
                            continue;
                        }
                        if(a[ni][nj]==1){
                            a[ni][nj] = 0;
                        }
                        else{
                            a[ni][nj] = 1;
                        }
                        sb.append((ni+1)+" "+(nj+1)+" ");
                    }
                    sb.append("\n");
                    for(int k = 0; k < 4; k++){
                        int ni = i+di[k];
                        int nj = j+dj[k];
                        if(a[ni][nj]==1){
                            sb.append((ni+1)+" "+(nj+1)+" ");
                        }
                    }
                    sb.append("\n");
                    for(int k = 0; k < 4; k++){
                        int ni = i+di[k];
                        int nj = j+dj[k];
                        a[ni][nj] = 0;
                    }
                }
            }
        }
        out.println(count);
        out.print(sb);
    }
    static class Node {
        ArrayList<Node> connect = new ArrayList<>();
        int comp = -1;

        public int gsize() {
            return connect.size();
        }
    }

//    static long nC2(long n) {
//        return add((n * (n + 1)) / 2, 0);
//    }
    public static int lowerBound(int key, int[] a) {
        int s = 0;
        int e = a.length - 1;
        if (e == -1) {
            return 0;
        }
        int ans = -1;
        while (s <= e) {
            int m = s + (e - s) / 2;
            if (a[m] >= key) {
                ans = m;
                e = m - 1;
            } else {
                s = m + 1;
            }
        }
        return ans == -1 ? s : ans;
    }

    public static int upperBound(int key, int[] a) {
        int s = 0;
        int e = a.length - 1;
        if (e == -1) {
            return 0;
        }
        int ans = -1;
        while (s <= e) {
            int m = s + (e - s) / 2;
            if (a[m] > key) {
                ans = m;
                e = m - 1;
            } else {
                s = m + 1;
            }
        }
        return ans == -1 ? s : ans;
    }

    public static long c2(long n) {
        if ((n & 1) == 0) {
            return mul(n / 2, n - 1);
        } else {
            return mul(n, (n - 1) / 2);
        }
    }

    public static long mul(long a, long b) {
        return ((a % MOD) * (b % MOD)) % MOD;
    }

    public static int add(int a, int b) {
        return ((a % MOD) + (b % MOD)) % MOD;
    }

    public static long sub(long a, long b) {
        return ((a % MOD) - (b % MOD)) % MOD;
    }

    ////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//Integer.lowestOneBit(i)   Equals k where k is the position of the first one in the binary
//Integer.highestOneBit(i)  Equals k where k is the position of the last one in the binary
//Integer.bitCount(i)       returns the number of one-bits
//Collections.sort(A,(p1,p2)->(int)(p2.x-p1.x)) To sort ArrayList in descending order wrt values of x.
//            Arrays.parallelSort(a,new Comparator<TPair>() {
//      		public int compare(TPair a,TPair b) {
//        			if(a.y==b.y) return a.x-b.x;
//        			return b.y-a.y;
//        		}
//        	});
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//PrimeFactors    
    public static ArrayList<Long> primeFactors(long n) {
        ArrayList<Long> arr = new ArrayList<>();
        if (n % 2 == 0)
            arr.add((long) 2);
        while (n % 2 == 0)
            n /= 2;
        for (long i = 3; i <= Math.sqrt(n); i += 2) {
            int flag = 0;
            while (n % i == 0) {
                n /= i;
                flag = 1;
            }
            if (flag == 1)
                arr.add(i);
        }
        if (n > 2)
            arr.add(n);
        return arr;
    }

    //Pair Class
    static class Pair implements Comparable<Pair> {
        int x;
        int y;

        public Pair(int x, int y) {
            this.x = x;
            this.y = y;
        }
        @Override
        public int compareTo(Pair o) {
            if (this.x == o.x) {
                return (this.y - o.y);
            }
            return this.x-o.x;
        }
    }

    static class TPair implements Comparable<TPair> {
        int l;
        int r;
        int val;

        public TPair(int l, int r, int index) {
            this.l = l;
            this.r = r;
            this.val = val;
        }

        @Override
        public int compareTo(TPair o) {
            if (this.l == o.l) {
                return (this.r - o.r);
            }
            return (this.l - o.l);
        }
    }

    //nCr
    static long[] facts, factInvs;

    static long exp(long base, long e) {
        if (e == 0) return 1;
        long half = exp(base, e / 2);
        if (e % 2 == 0) return mul(half, half);
        return mul(half, mul(half, base));
    }

    static long modInv(long x) {
        return exp(x, MOD - 2);
    }

    static void precomp() {
        facts = new long[1_000_000];
        factInvs = new long[1_000_000];
        factInvs[0] = facts[0] = 1;
        for (int i = 1; i < facts.length; i++)
            facts[i] = mul(facts[i - 1], i);
        factInvs[facts.length - 1] = modInv(facts[facts.length - 1]);
        for (int i = facts.length - 2; i >= 0; i--)
            factInvs[i] = mul(factInvs[i + 1], i + 1);
    }

    static long nCk(int n, int k) {
        return mul(facts[n], mul(factInvs[k], factInvs[n - k]));
    }

    //Kadane's Algorithm
    static long maxSubArraySum(ArrayList<Long> a) {
        if (a.size() == 0) {
            return 0;
        }
        long max_so_far = a.get(0);
        long curr_max = a.get(0);
        for (int i = 1; i < a.size(); i++) {
            curr_max = Math.max(a.get(i), curr_max + a.get(i));
            max_so_far = Math.max(max_so_far, curr_max);
        }
        return max_so_far;
    }

    //Shuffle Sort
    static final Random random = new Random();

    static void ruffleSort(int[] a) {
        int n = a.length;//shuffle, then sort
        for (int i = 0; i < n; i++) {
            int oi = random.nextInt(n), temp = a[oi];
            a[oi] = a[i];
            a[i] = temp;
        }
        Arrays.sort(a);
    }

    //Merge Sort
    static void merge(long arr[], int l, int m, int r) {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;

        /* Create temp arrays */
        long L[] = new long[n1];
        long R[] = new long[n2];

        /*Copy data to temp arrays*/
        for (int i = 0; i < n1; ++i)
            L[i] = arr[l + i];
        for (int j = 0; j < n2; ++j)
            R[j] = arr[m + 1 + j];


        /* Merge the temp arrays */

        // Initial indexes of first and second subarrays
        int i = 0, j = 0;

        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2) {
            if (L[i] <= R[j]) {
                arr[k] = L[i];
                i++;
            } else {
                arr[k] = R[j];
                j++;
            }
            k++;
        }

        /* Copy remaining elements of L[] if any */
        while (i < n1) {
            arr[k] = L[i];
            i++;
            k++;
        }

        /* Copy remaining elements of R[] if any */
        while (j < n2) {
            arr[k] = R[j];
            j++;
            k++;
        }
    }

    // Main function that sorts arr[l..r] using
    // merge()
    static void sort(long arr[], int l, int r) {
        if (l < r) {
            // Find the middle point
            int m = (l + r) / 2;

            // Sort first and second halves
            sort(arr, l, m);
            sort(arr, m + 1, r);

            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }

    //Brian Kernighans Algorithm
    static long countSetBits(long n) {
        if (n == 0) return 0;
        return 1 + countSetBits(n & (n - 1));
    }

    //Factorial
    static long factorial(long n) {
        if (n == 1) return 1;
        if (n == 2) return 2;
        if (n == 3) return 6;
        return n * factorial(n - 1);
    }

    //Euclidean Algorithm
    static long gcd(long A, long B) {
        if (B == 0) return A;
        return gcd(B, A % B);
    }

    //Modular Exponentiation
    static long fastExpo(long x, long n) {
        if (n == 0) return 1;
        if ((n & 1) == 0) return fastExpo((x * x) % MOD, n / 2) % MOD;
        return ((x % MOD) * fastExpo((x * x) % MOD, (n - 1) / 2)) % MOD;
    }

    //AKS Algorithm
    static boolean isPrime(long n) {
        if (n <= 1) return false;
        if (n <= 3) return true;
        if (n % 2 == 0 || n % 3 == 0) return false;
        for (int i = 5; i <= Math.sqrt(n); i += 6)
            if (n % i == 0 || n % (i + 2) == 0) return false;
        return true;
    }

    //Reverse an array
    static <T> void reverse(T arr[], int l, int r) {
        Collections.reverse(Arrays.asList(arr).subList(l, r));
    }

    //Print array
    static void print1d(long arr[]) {
        out.println(Arrays.toString(arr));
    }

    static void print2d(int arr[][]) {
        for (int a[] : arr) out.println(Arrays.toString(a));
    }

    //Sieve of eratosthenes
    static int[] findPrimes(int n) {
        boolean isPrime[] = new boolean[n + 1];
        ArrayList<Integer> a = new ArrayList<>();
        int result[];
        Arrays.fill(isPrime, true);
        isPrime[0] = false;
        isPrime[1] = false;
        for (int i = 2; i * i <= n; ++i) {
            if (isPrime[i] == true) {
                for (int j = i * i; j <= n; j += i) isPrime[j] = false;
            }
        }
        for (int i = 0; i <= n; i++) if (isPrime[i] == true) a.add(i);
        result = new int[a.size()];
        for (int i = 0; i < a.size(); i++) result[i] = a.get(i);
        return result;

    }

    //Indivisual factors of all nos till n
    static ArrayList<Integer>[] indiFactors(int n) {
        ArrayList<Integer>[] A = new ArrayList[n + 1];
        for (int i = 0; i <= n; i++) {
            A[i] = new ArrayList<>();
        }
        int[] sieve = new int[n + 1];
        for (int i = 2; i <= n; i++) {
            if (sieve[i] == 0) {
                for (int j = i; j <= n; j += i)
                    if (sieve[j] == 0) {
                        //sieve[j]=i;
                        A[j].add(i);
                    }
            }
        }
        return A;
    }

    //Segmented Sieve
    static boolean[] segmentedSieve(long l, long r) {
        boolean[] segSieve = new boolean[(int) (r - l + 1)];
        Arrays.fill(segSieve, true);
        int[] prePrimes = findPrimes((int) Math.sqrt(r));
        for (int p : prePrimes) {
            long low = (l / p) * p;
            if (low < l) {
                low += p;
            }
            if (low == p) {
                low += p;
            }
            for (long j = low; j <= r; j += p) {
                segSieve[(int) (j - l)] = false;
            }
        }
        if (l == 1) {
            segSieve[0] = false;
        }
        return segSieve;
    }

    //Euler Totent function
    static long countCoprimes(long n) {
        ArrayList<Long> prime_factors = new ArrayList<>();
        long x = n, flag = 0;
        while (x % 2 == 0) {
            if (flag == 0) prime_factors.add(2L);
            flag = 1;
            x /= 2;
        }
        for (long i = 3; i * i <= x; i += 2) {
            flag = 0;
            while (x % i == 0) {
                if (flag == 0) prime_factors.add(i);
                flag = 1;
                x /= i;
            }
        }
        if (x > 2) prime_factors.add(x);
        double ans = (double) n;
        for (Long p : prime_factors) {
            ans *= (1.0 - (Double) 1.0 / p);
        }
        return (long) ans;
    }

    static long modulo = (long) 1e9 + 7;

    public static long modinv(long x) {
        return modpow(x, modulo - 2);
    }

    public static long modpow(long a, long b) {
        if (b == 0) {
            return 1;
        }
        long x = modpow(a, b / 2);
        x = (x * x) % modulo;
        if (b % 2 == 1) {
            return (x * a) % modulo;
        }
        return x;
    }

    public static class FastReader {
        BufferedReader br;
        StringTokenizer st;
        //it reads the data about the specified point and divide the data about it ,it is quite fast
        //than using direct

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (Exception r) {
                    r.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());//converts string to integer
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (Exception r) {
                r.printStackTrace();
            }
            return str;
        }
    }
}