A star is a figure of the following type: an asterisk character '*' in the center of the figure and four rays (to the left, right, top, bottom) of the same positive length. The size of a star is the length of its rays. The size of a star must be a positive number (i.e. rays of length 0 are not allowed).

Let's consider empty cells are denoted by '.', then the following figures are stars:

<image> The leftmost figure is a star of size 1, the middle figure is a star of size 2 and the rightmost figure is a star of size 3.

You are given a rectangular grid of size n × m consisting only of asterisks '*' and periods (dots) '.'. Rows are numbered from 1 to n, columns are numbered from 1 to m. Your task is to draw this grid using any number of stars or find out that it is impossible. Stars can intersect, overlap or even coincide with each other. The number of stars in the output can't exceed n ⋅ m. Each star should be completely inside the grid. You can use stars of same and arbitrary sizes.

In this problem, you do not need to minimize the number of stars. Just find any way to draw the given grid with at most n ⋅ m stars.

Input

The first line of the input contains two integers n and m (3 ≤ n, m ≤ 100) — the sizes of the given grid.

The next n lines contains m characters each, the i-th line describes the i-th row of the grid. It is guaranteed that grid consists of characters '*' and '.' only.

Output

If it is impossible to draw the given grid using stars only, print "-1".

Otherwise in the first line print one integer k (0 ≤ k ≤ n ⋅ m) — the number of stars needed to draw the given grid. The next k lines should contain three integers each — x_j, y_j and s_j, where x_j is the row index of the central star character, y_j is the column index of the central star character and s_j is the size of the star. Each star should be completely inside the grid.

Examples

Input

6 8
....*...
...**...
..*****.
...**...
....*...
........


Output

3
3 4 1
3 5 2
3 5 1


Input

5 5
.*...
****.
.****
..**.
.....


Output

3
2 2 1
3 3 1
3 4 1


Input

5 5
.*...
***..
.*...
.*...
.....


Output

-1


Input

3 3
*.*
.*.
*.*


Output

-1

Note

In the first example the output 
    
    
    2  
    3 4 1  
    3 5 2  
    

is also correct.
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.Scanner; 
import java.util.StringTokenizer; 
import java.util.*;
import java.io.*;
public class codeforces 
{ 
   static class Student{
       int x,y;
        Student(int x,int y){
            this.x=x;
            this.y=y;
            //this.z=z;
        }
    }
    static int prime[];
    static void sieveOfEratosthenes(int n) 
    { 
        // Create a boolean array "prime[0..n]" and initialize 
        // all entries it as true. A value in prime[i] will 
        // finally be false if i is Not a prime, else true. 
        int pos=0;
        prime= new int[n+1]; 
        for(int p = 2; p*p <=n; p++) 
        { 
            // If prime[p] is not changed, then it is a prime 
            if(prime[p] == 0)
            { 
                // Update all multiples of p
                prime[p]=p;
                for(int i = p*p; i <= n; i += p)
                    if(prime[i]==0)
                    prime[i] = p; 
            } 
        } 
    }
static    class Sortbyroll implements Comparator<Student> 
{ 
    // Used for sorting in ascending order of 
    // roll number 
    public int compare(Student c, Student b) 
    { 
        return c.y-b.y;
    } 
} 
static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    } 
 
    static class Edge{
        int a,b;
        Edge(int a,int b){
            this.a=a;
            this.b=b;
        }
    }
    static class Trie{
        HashMap<Character,Trie>map;
         int c;
        Trie(){
            map=new HashMap<>();
            //z=null;
            //o=null;
            c=0;
        }
    }
    //static long ans;
    static int parent[];
    static int rank[];
   // static int b[][];
    static int bo[];
    static int ho[];
    static int seg[];
    //static int pos;
     static long mod=1000000007;
    //static int dp[][];
    static HashMap<Integer,Integer>map;
    static PriorityQueue<Student>q=new PriorityQueue<>();
    //static Stack<Integer>st;
   // static ArrayList<Character>ans;
    static ArrayList<ArrayList<Integer>>adj;
    //static long ans;
    static int pos;
    static Trie root;
    static long fac[];
    static int gw,gb;
    //static long mod=(long)(998244353);
    //static int ans;
    static void solve()throws IOException{
        FastReader sc=new FastReader();
        int n,m,i,j,k,pos=0,min,ans=0;
        StringBuilder sb=new StringBuilder();
        n=sc.nextInt();
        m=sc.nextInt();
        String s[]=new String[n];
        for(i=0;i<n;i++)
            s[i]=sc.nextLine();
        int a[][]=new int[n][m];
        for(i=0;i<n;i++){
            for(j=0;j<m;j++){
                if(s[i].charAt(j)!='*')
                    continue;
                min=Integer.MAX_VALUE;
                k=j;
                while(k<m&&s[i].charAt(k)=='*')
                    ++k;
                min=Math.min(min,k-j-1);
                k=j;
                while(k>=0&&s[i].charAt(k)=='*')
                    --k;
                min=Math.min(min,j-k-1);
                k=i;
                while(k<n&&s[k].charAt(j)=='*')
                    ++k;
                min=Math.min(min,k-i-1);
                k=i;
                while(k>=0&&s[k].charAt(j)=='*')
                    --k;
                min=Math.min(min,i-k-1);
                //min-=1;
                if(min==0)
                    continue;
                a[i][j]=1;
                ++ans;
                sb.append((i+1)+" "+(j+1)+" "+min+"\n");
                k=1;
                //System.out.println(min+" "+i+" "+j);
                while(k<=min){
                    a[i+k][j]=1;
                    a[i-k][j]=1;
                    a[i][j+k]=1;
                    a[i][j-k]=1;
                    ++k;
                }
            }
        }
        for(i=0;i<n;i++){
            for(j=0;j<m;j++){
                if(s[i].charAt(j)!='*')
                    continue;
                if(a[i][j]!=1){
                    pos=1;
                    break;
                }
            }
            if(pos==1)
                break;
        }
        if(pos==1)
            System.out.println(-1);
        else{
            System.out.println(ans);
            System.out.println(sb.toString());
        }
    }
    static long fac(long x,long y){
        if(y==0||x==0)
            return 1;
        return x * fac(x-1,y-1);
    }
    static long nCr(long n, long r, 
                                    long p) 
    { 
        return (fac[(int)n]* modInverse(fac[(int)r], p) 
                % p * modInverse(fac[(int)(n-r)], p) 
                                    % p) % p; 
    } 
    //static int prime[];
    //static int dp[];
    public static void main(String[] args){
        //long sum=0;
        try {
                codeforces.solve();
                } catch (Exception e) {
                    e.printStackTrace();
                }
    }
       static long modInverse(long n, long p) 
    { 
        return power(n, p-(long)2,p); 
    } 
 static long power(long x, long y, long p) 
{ 
    long res = 1;      // Initialize result 
  
    x = x % p;  // Update x if it is more than or  
                // equal to p 
  
    while (y > 0) 
    { 
        // If y is odd, multiply x with result 
        if (y %(long)2!=0) 
            res = (res*x) % p; 
  
        // y must be even now 
        y = y>>1; // y = y/2 
        x = (x*x) % p;   
    } 
    return res%p;
 }
    /*public static long power(long x,long a) {
		if(a==0) return 1;
		if(a%2==1)return (x*1L*power(x,a-1))%mod;
		return (power((int)((x*1L*x)%mod),a/2))%mod;
	}*/
   static int find(int x) 
    { 
        // Finds the representative of the set 
        // that x is an element of 
        while(parent[x]!=x) 
        { 
            // if x is not the parent of itself 
            // Then x is not the representative of 
            // his set,
            x=parent[x];
  
            // so we recursively call Find on its parent 
            // and move i's node directly under the 
            // representative of this set 
        } 
  
        return x; 
    }
    static void union(int x, int y) 
    { 
        // Find representatives of two sets 
        int xRoot = find(x), yRoot = find(y); 
  
        // Elements are in the same set, no need 
        // to unite anything. 
        if (xRoot == yRoot) 
            return; 
  
         // If x's rank is less than y's rank 
        if (rank[xRoot] < rank[yRoot]) 
  
            // Then move x under y  so that depth 
            // of tree remains less 
            parent[xRoot] = yRoot; 
  
        // Else if y's rank is less than x's rank 
        else if (rank[yRoot] < rank[xRoot]) 
  
            // Then move y under x so that depth of 
            // tree remains less 
            parent[yRoot] = xRoot; 
  
        else // if ranks are the same 
        { 
            // Then move y under x (doesn't matter 
            // which one goes where) 
            parent[yRoot] = xRoot; 
  
            // And increment the the result tree's 
            // rank by 1 
            rank[xRoot] = rank[xRoot] + 1; 
        } 
    } 
    static long gcd(long a, long b) 
    { 
        if (a == 0){ 
             //ans+=b;
            return b;
        }
        return gcd(b % a, a); 
    }
}