You are given a rectangular field of n × m cells. Each cell is either empty or impassable (contains an obstacle). Empty cells are marked with '.', impassable cells are marked with '*'. Let's call two empty cells adjacent if they share a side.

Let's call a connected component any non-extendible set of cells such that any two of them are connected by the path of adjacent cells. It is a typical well-known definition of a connected component.

For each impassable cell (x, y) imagine that it is an empty cell (all other cells remain unchanged) and find the size (the number of cells) of the connected component which contains (x, y). You should do it for each impassable cell independently.

The answer should be printed as a matrix with n rows and m columns. The j-th symbol of the i-th row should be "." if the cell is empty at the start. Otherwise the j-th symbol of the i-th row should contain the only digit —- the answer modulo 10. The matrix should be printed without any spaces.

To make your output faster it is recommended to build the output as an array of n strings having length m and print it as a sequence of lines. It will be much faster than writing character-by-character.

As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 1000) — the number of rows and columns in the field.

Each of the next n lines contains m symbols: "." for empty cells, "*" for impassable cells.

Output

Print the answer as a matrix as described above. See the examples to precise the format of the output.

Examples

Input

3 3
*.*
.*.
*.*


Output

3.3
.5.
3.3


Input

4 5
**..*
..***
.*.*.
*.*.*


Output

46..3
..732
.6.4.
5.4.3

Note

In first example, if we imagine that the central cell is empty then it will be included to component of size 5 (cross). If any of the corner cell will be empty then it will be included to component of size 3 (corner).
import java.util.*;
import java.lang.*;
import java.io.*;
public class Codechef
{       static PrintWriter out=new PrintWriter(System.out);static FastScanner in = new FastScanner(System.in);static class FastScanner {BufferedReader br;StringTokenizer stok;FastScanner(InputStream is) {br = new BufferedReader(new InputStreamReader(is));}
        String next() throws IOException {while (stok == null || !stok.hasMoreTokens()) {String s = br.readLine();if (s == null) {return null;}
                stok = new StringTokenizer(s);}return stok.nextToken();}
        int ni() throws IOException {return Integer.parseInt(next());}long nl() throws IOException {return Long.parseLong(next());}double nd() throws IOException {return Double.parseDouble(next());}char nc() throws IOException {return (char) (br.read());}String ns() throws IOException {return br.readLine();}
        int[] nia(int n) throws IOException{int a[] = new int[n];for (int i = 0; i < n; i++)a[i] = ni();return a;}long[] nla(int n) throws IOException {long a[] = new long[n];for (int i = 0; i < n; i++)a[i] = nl();return a;}
        double[] nda(int n)throws IOException {double a[] = new double[n];for (int i = 0; i < n; i++)a[i] = nd();return a;}int [][] imat(int n,int m) throws IOException{int mat[][]=new int[n][m];for(int i=0;i<n;i++){for(int j=0;j<m;j++)mat[i][j]=ni();}return mat;}
    }
	static long mod=Long.MAX_VALUE;
	static boolean visited[][];
	static int a[][],temp[][];
	static ArrayList<Integer> arr[];
	static int dir=0,n,m,ans,counter=0,length=0;
	public static void main (String[] args) throws java.lang.Exception
	{   int i,j;
	    HashMap<Integer,Integer> hm=new HashMap<Integer,Integer>();
		/*  if(hm.containsKey(z))
		        hm.put(z,hm.get(z)+1);
		    else
		        hm.put(z,1);
		 */       
        n=in.ni();
        m=in.ni();
        a=new int[n][m];
        for(i=0;i<n;i++)
        {   String s=in.next();
            for(j=0;j<m;j++)
               if(s.charAt(j)=='*')
                    a[i][j]=1;
        }
        temp=new int[n][m];
        int temp2[]=new int[1000040]; 
        visited=new boolean[n][m];
        for(i=0;i<n;i++)
        {   for(j=0;j<m;j++)
            {   if(!visited[i][j] && a[i][j]==0)
                {   counter++;
                    length=0;
                    dfs(i,j);
                    temp2[counter]=length;
                }
            }
        }
        for(i=0;i<n;i++)
        {   for(j=0;j<m;j++)
            {   if(a[i][j]==0)
                    out.print(".");
                else
                {   int p=0,q=0,r=0,s=0;
                    HashSet<Integer> set2=new HashSet<Integer>();
                    if(j+1<m && a[i][j+1]==0)
    	               set2.add(temp[i][j+1]);
    	            if(i+1<n && a[i+1][j]==0)
    	               set2.add(temp[i+1][j]);
                	if(j-1>=0 && a[i][j-1]==0)
    	               set2.add(temp[i][j-1]);
    	            if(i-1>=0 && a[i-1][j]==0)
    	               set2.add(temp[i-1][j]);
    	            int z=1;
    	            for(int c:set2)
    	                z+=temp2[c];
                    z=z%10;
                    out.print(z);
                }
            }
            out.println();
        }
        out.close();
	}
	static void dfs(int i,int j)
	{   if(visited[i][j])
    	    return;
    	visited[i][j]=true;
    	temp[i][j]=counter;
    	length++;
    	if(j+1<m && a[i][j+1]==0)
    	       dfs(i,j+1);
    	if(i+1<n && a[i+1][j]==0)
    	       dfs(i+1,j);
    	if(j-1>=0 && a[i][j-1]==0)
    	       dfs(i,j-1);
    	if(i-1>=0 && a[i-1][j]==0)
    	       dfs(i-1,j);
    	return ;
	}
	static class pair implements Comparable<pair>{
		int x, y;
		public pair(int x, int y){this.x = x; this.y = y;}
		@Override
		public int compareTo(pair arg0) 
		{   if(x<arg0.x)    return -1;
		    else if(x==arg0.x)
		    {   if(y<arg0.y)    return -1;
		        else if(y>arg0.y)   return 1;
		        else    return 0;
		    }
		    else    return 1;
		}
	}
	static long gcd(long a,long b)
	{   if(b==0)
	        return a;
	    return gcd(b,a%b);    
	}
	static long exponent(long a,long n)
	{   long ans=1;
	    while(n!=0)
	    {   if(n%2==1)
	            ans=(ans*a)%mod;
	       a=(a*a)%mod;
	       n=n>>1;
	    }
	    return ans;
	}
	static int binarySearch(int a[], int item, int low, int high) 
    {   if (high <= low) 
            return (item > a[low])?  (low + 1): low; 
        int mid = (low + high)/2; 
        if(item == a[mid]) 
            return mid+1; 
        if(item > a[mid]) 
            return binarySearch(a, item, mid+1, high); 
        return binarySearch(a, item, low, mid-1); 
    } 
    static void merge(int arr[], int l, int m, int r) {   
        int n1 = m - l + 1; int n2 = r - m; int L[] = new int [n1]; int R[] = new int [n2]; 
        
        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]; int i = 0, j = 0; 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++; } while (i < n1){   arr[k] = L[i]; i++; k++; } while (j < n2) { arr[k] = R[j];   j++;     k++; } 
    } 
    static void Sort(int arr[], int l, int r) {if (l < r) {   int m = (l+r)/2; Sort(arr, l, m); Sort(arr , m+1, r); merge(arr, l, m, r); } } 
    
    static void sort(int a[])
    {Sort(a,0,a.length-1);}
}