Polycarp and his friends want to visit a new restaurant. The restaurant has n tables arranged along a straight line. People are already sitting at some tables. The tables are numbered from 1 to n in the order from left to right. The state of the restaurant is described by a string of length n which contains characters "1" (the table is occupied) and "0" (the table is empty).

Restaurant rules prohibit people to sit at a distance of k or less from each other. That is, if a person sits at the table number i, then all tables with numbers from i-k to i+k (except for the i-th) should be free. In other words, the absolute difference of the numbers of any two occupied tables must be strictly greater than k.

For example, if n=8 and k=2, then:

  * strings "10010001", "10000010", "00000000", "00100000" satisfy the rules of the restaurant; 
  * strings "10100100", "10011001", "11111111" do not satisfy to the rules of the restaurant, since each of them has a pair of "1" with a distance less than or equal to k=2. 



In particular, if the state of the restaurant is described by a string without "1" or a string with one "1", then the requirement of the restaurant is satisfied.

You are given a binary string s that describes the current state of the restaurant. It is guaranteed that the rules of the restaurant are satisfied for the string s.

Find the maximum number of free tables that you can occupy so as not to violate the rules of the restaurant. Formally, what is the maximum number of "0" that can be replaced by "1" such that the requirement will still be satisfied?

For example, if n=6, k=1, s= "100010", then the answer to the problem will be 1, since only the table at position 3 can be occupied such that the rules are still satisfied.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases in the test. Then t test cases follow.

Each test case starts with a line containing two integers n and k (1 ≤ k ≤ n ≤ 2⋅ 10^5) — the number of tables in the restaurant and the minimum allowed distance between two people.

The second line of each test case contains a binary string s of length n consisting of "0" and "1" — a description of the free and occupied tables in the restaurant. The given string satisfy to the rules of the restaurant — the difference between indices of any two "1" is more than k.

The sum of n for all test cases in one test does not exceed 2⋅ 10^5.

Output

For each test case output one integer — the number of tables that you can occupy so as not to violate the rules of the restaurant. If additional tables cannot be taken, then, obviously, you need to output 0.

Example

Input


6
6 1
100010
6 2
000000
5 1
10101
3 1
001
2 2
00
1 1
0


Output


1
2
0
1
1
1

Note

The first test case is explained in the statement.

In the second test case, the answer is 2, since you can choose the first and the sixth table.

In the third test case, you cannot take any free table without violating the rules of the restaurant.
import java.io.*;
import java.math.BigDecimal;
import java.util.*;

public class scratch_25 {
    static class Reader {
        static BufferedReader reader;
        static StringTokenizer tokenizer;

        /** call this method to initialize reader for InputStream */
        static void init(InputStream input) {
            reader = new BufferedReader(
                    new InputStreamReader(input) );
            tokenizer = new StringTokenizer("");
        }

        /** get next word */
        static String next() throws IOException {
            while ( ! tokenizer.hasMoreTokens() ) {
                //TODO add check for eof if necessary
                tokenizer = new StringTokenizer(
                        reader.readLine() );
            }
            return tokenizer.nextToken();
        }

        static int nextInt() throws IOException {
            return Integer.parseInt( next() );
        }

        static double nextDouble() throws IOException {
            return Double.parseDouble( next() );
        }
        static long nextLong() throws IOException{
            return Long.parseLong(next());
        }
    }
    static class pair
    {
        long first, second;
        public pair(long first, long second)
        {
            this.first = first;
            this.second = second;
        }
    }
    // To find the power of a prime p in
// factorial N
    static long findPowerOfP(long n, long p)
    {
        long count = 0;
        long r = p;
        while (n>0)
        {
            n/=r;

            // calculating Math.floor(n/r)
            // and adding to the count
            count += n ;

            // increasing the power of p
            // from 1 to 2 to 3 and so on

        }
        return count;
    }

    // returns all the prime factors of k

    // Returns largest power of k that
// divides n!
    static long largestPowerOfK(long n, long k)
    {
        Vector<pair > vec = new Vector<pair>();
        vec = primeFactorsofK(k);
        long ans = Long.MAX_VALUE;
        for (int i = 0; i < vec.size(); i++){
          //  System.out.println("bhosdike");

            // calculating minimum power of all
            // the prime factors of k
            ans = Math.min(ans, findPowerOfP(n,
                    vec.get(i).first) / vec.get(i).second);}

        return ans;
    }
    public static Vector<pair> primeFactorsofK(long n)
    {
        Vector<pair> ans = new Vector<pair>();
        // Print the number of 2s that divide n
        long count=0;
        while (n%2==0)
        {
            count++;
           // System.out.print(2 + " ");
            n /= 2;
        }
        if(count>0){
        ans.add(new pair(2,count));}

        // n must be odd at this point.  So we can
        // skip one element (Note i = i +2)
        for (int i = 3; i <= Math.sqrt(n); i+= 2)
        {
            long count1=0;
            // While i divides n, print i and divide n
            while (n%i == 0)
            {
                count1++;
               // System.out.print(i + " ");
                n /= i;
            }
            if(count1>0){
            ans.add(new pair(i,count1));}
        }

        // This condition is to handle the case whien
        // n is a prime number greater than 2
        if(n>2){
            ans.add(new pair(n,1));
        }
//

        return ans;
    }
    static int longSubarrWthSumDivByK(int arr[],
                                      int n, int k)
    {
        // unodered map 'um' implemented as
        // hash table
        HashMap<Integer, Integer> um= new HashMap<Integer, Integer>();

        // 'mod_arr[i]' stores (sum[0..i] % k)
        int mod_arr[]= new int[n];
        int max = 0;
        int curr_sum = 0;

        // traverse arr[] and build up the
        // array 'mod_arr[]'
        for (int i = 0; i < n; i++)
        {
            curr_sum += arr[i];

            // as the sum can be negative,
            // taking modulo twice
            mod_arr[i] = ((curr_sum % k) + k) % k;
        }

        for (int i = 0; i < n; i++)
        {
            // if true then sum(0..i) is
            // divisible by k
            if (mod_arr[i] == 0)
                // update 'max'
                max = i + 1;

                // if value 'mod_arr[i]' not present in 'um'
                // then store it in 'um' with index of its
                // first occurrence
            else if (um.containsKey(mod_arr[i]) == false)
                um.put(mod_arr[i] , i);

            else
                // if true, then update 'max'
                if (max < (i - um.get(mod_arr[i])))
                    max = i - um.get(mod_arr[i]);
        }

        // required length of longest subarray with
        // sum divisible by 'k'
        return max;
    }

    public static void main(String[] args) throws IOException {
        Reader.init(System.in);
        int t= Reader.nextInt();
        for (int j = 0; j <t ; j++) {
            int n= Reader.nextInt();
            int k= Reader.nextInt();
            String str= Reader.next();
            int count=0;
            int f= str.length();
            for (int i = 0; i <f ; i++) {
                if(str.charAt(i)=='1'){
                    count++;
                }
            }
            int arr1[]= new int[count];
            int counter=0;
            for (int i = 0; i <f ; i++) {
                if(str.charAt(i)=='1'){
                    arr1[counter]=i;
                    counter++;
                }
            }
           // System.out.println(Arrays.toString(arr1));
            int ans=0;
           // System.out.println(count);
           // System.out.println(n/k);
            if(count==0){
                int l=0;
              while(l<n){
                  ans+=1;
                  l+=k+1;
              }
              if(ans==0){
                  System.out.println(1);
              }
              else{
                  System.out.println(ans);
              }
            }

            else if(count!=0){
            int s=arr1[0];   // index of first 1
            if(s>k){
                s-=k;
                int l=0;
                while(l<s){
                    ans+=1;
                    l+=k+1;
                }



            }
            int e= arr1[count-1];// index of last 1
            if(n-e-1>k){
                int g= n-e-1;
                g-=k;
                int u=0;
                while(u<g){
                    ans+=1;
                    u+=k+1;
                }
               // ans+=Math.ceil((1.0*l)/(1.0*k));
            }

            for (int i = 0; i <count-1 ; i++) {
                int st= arr1[i];
                int ed= arr1[i+1];
                int all= ed-st-1;
                all-=2*k;
                int p=0;
                while(p<all){
                    ans+=1;
                    p+=k+1;
                }





            }


            System.out.println(ans);}



        }

    }
    public static boolean func(int arr[], int mid, int x,int n){
     //   System.out.println("mid="+mid);
        int sum=0;
        boolean b=false;
        for (int i = 0; i <mid ; i++) {
            sum+=arr[i];
        }
     //   System.out.println(sum);
        if(sum%x==0){
            return false;
        }
        for (int i = 1; i <n-mid+1 ; i++) {
            sum-=arr[i-1];
            sum+=arr[i+mid-1];
            //System.out.println(sum);
            if(sum%x==0){
                return false;
            }

        }
        return true;






    }



    static void merge(int 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 */
        int L[] = new int [n1];
        int R[] = new int [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(int 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);
        }
    }
    static boolean ans(int check, long arr[],int n,int med,int k){
        long ans=0;
        for (int i = med; i <n ; i++) {
            if(check>=arr[i]){
                ans+=check-arr[i];
            }
        }
        return ans<=k;




    }

    static long smallestDivisor(long n)
    {
        // if divisible by 2
        if (n % 2 == 0)
            return 2;

        // iterate from 3 to sqrt(n)
        for (int i = 3; (long)i *(long) i <= n; i += 2) {
            if (n % i == 0)
                return i;
        }

        return n;
    }
    static boolean isPrime(long n)
    {
        // Corner cases
        if (n <= 1) return false;
        if (n <= 3) return true;

        // This is checked so that we can skip
        // middle five numbers in below loop
        if (n % 2 == 0 || n % 3 == 0) return false;

        for (long i = 5; i * i <= n; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;

        return true;
    }
static BigDecimal fd(int mid, double h, double c){
        BigDecimal a = new BigDecimal((h+c)/(2.0));
        BigDecimal b = new BigDecimal((h-c)/(2.0*mid));
        BigDecimal cd = a.add(b);
        return cd;




}
    static   int partition(double arr[],int arr1[],int arr2[], int low, int high)
    {
        double pivot = arr[high];
        int i = (low-1); // index of smaller element
        for (int j=low; j<high; j++)
        {
            // If current element is smaller than the pivot
            if (arr[j] < pivot)
            {
                i++;

                // swap arr[i] and arr[j]
                double temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
                int temp1=arr1[i];
                arr1[i]=arr1[j];
                arr1[j]=temp1;
                int temp2=arr2[i];
                arr2[i]=arr2[j];
                arr2[j]=temp2;

            }
        }

        // swap arr[i+1] and arr[high] (or pivot)
        double temp = arr[i+1];
        arr[i+1] = arr[high];
        arr[high] = temp;
        int temp1= arr1[i+1];
        arr1[i+1]=arr1[high];
        arr1[high]= temp1;
        int temp2= arr2[i+1];
        arr2[i+1]=arr2[high];
        arr2[high]= temp2;

        return i+1;
    }


    /* The main function that implements QuickSort()
      arr[] --> Array to be sorted,
      low  --> Starting index,
      high  --> Ending index */
    static void sort(double arr[],int arr1[],int arr2[], int low, int high)
    {
        if (low < high)
        {
            /* pi is partitioning index, arr[pi] is
              now at right place */
            int pi = partition(arr,arr1,arr2, low, high);

            // Recursively sort elements before
            // partition and after partition
            sort(arr,arr1,arr2, low, pi-1);
            sort(arr,arr1, arr2,pi+1, high);
        }
    }
    static int removeDuplicates(int arr[], int n)
    {
        // Return, if array is empty
        // or contains a single element
        if (n==0 || n==1)
            return n;

        int[] temp = new int[n];

        // Start traversing elements
        int j = 0;
        for (int i=0; i<n-1; i++)
            // If current element is not equal
            // to next element then store that
            // current element
            if (arr[i] != arr[i+1])
                temp[j++] = arr[i];

        // Store the last element as whether
        // it is unique or repeated, it hasn't
        // stored previously
        temp[j++] = arr[n-1];

        // Modify original array
        for (int i=0; i<j; i++)
            arr[i] = temp[i];

        return j;
    }
    static long maxPrimeFactors( long n)
    {
        // Initialize the maximum prime
        // factor variable with the
        // lowest one
        long maxPrime = -1;

        // Print the number of 2s
        // that divide n
        while (n % 2 == 0) {
            maxPrime = 2;

            // equivalent to n /= 2
            n >>= 1;
        }

        // n must be odd at this point,
        // thus skip the even numbers
        // and iterate only for odd
        // integers
        for (int i = 3; i <= Math.sqrt(n); i += 2) {
            while (n % i == 0) {
                maxPrime = i;
                n = n / i;
            }
        }

        // This condition is to handle
        // the case when n is a prime
        // number greater than 2
        if (n > 2)
            maxPrime = n;

        return maxPrime;
    }




}