You are given an H × W grid.
The squares in the grid are described by H strings, S_1,...,S_H.
The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W).
`.` stands for an empty square, and `#` stands for a square containing a bomb.

Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square.
(Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.)
He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square.

Print the strings after the process.

Constraints

* 1 \leq H,W \leq 50
* S_i is a string of length W consisting of `#` and `.`.

Input

Input is given from Standard Input in the following format:


H W
S_1
:
S_H


Output

Print the H strings after the process.
The i-th line should contain a string T_i of length W, where the j-th character in T_i corresponds to the square at the i-th row from the top and j-th row from the left in the grid (1 \leq i \leq H, 1 \leq j \leq W).

Examples

Input

3 5
.....
.#.#.
.....


Output

11211
1#2#1
11211


Input

3 5


Output




Input

6 6
.
.#.##
.#
.#..#.
.##..
.#...


Output

3
8#7##
5#
4#65#2
5##21
4#310
#include <bits/stdc++.h>
using namespace std;

#define inf 1072114514
#define llinf 4154118101919364364
#define mod 1000000007
#define pi 3.1415926535897932384


int round(int a,int b){if((a%b)*2 >= b){return (a/b)+1;}return a/b;} 
int gcd(int a,int b){int c;while(b!=0){c=a%b;a=b;b=c;}return a;}    //最大公約数
int lcm(int a,int b){int c=gcd(a,b);a/=c;return a*b;}     //最小公倍数
int nCr(int a,int b){int i,r=1;for(i=1;i<=b;i++){r*=(a+1-i);r/=i;}return r;} //コンビネーション
int nHr(int a,int b){return nCr(a+b-1,b);}       // 重複組み合わせ
int fact(int a){int i,r=1;for(i=1;i<=a;i++){r*=i;}return r;}  //階乗
int pow(int a,int b){int i,r=1;for(i=1;i<=b;i++){r*=a;}return r;} // a~bまでの階乗
int dsum(int x){int r=0;while(x){r+=(x%10);x/=10;}return r;}   //数字の各位の和
int dsumb(int x,int b){int r=0;while(x){r+=(x%b);x/=b;}return r;}  // b進数の各位の和？
int sankaku(int x){return ((1+x)*x)/2;} //三角数 xまでの和

//以下long long ver
long long llmax(long long a,long long b){if(a>b){return a;}return b;}
long long llmin(long long a,long long b){if(a<b){return a;}return b;}
long long llzt(long long a,long long b){return llmax(a,b)-llmin(a,b);}
long long llround(long long a,long long b){if((a%b)*2 >= b){return (a/b)+1;}return a/b;}
long long llceil(long long a,long long b){if(a%b==0){return a/b;}return (a/b)+1;}
long long llgcd(long long a,long long b){long long c;while(b!=0){c=a%b;a=b;b=c;}return a;}
long long lllcm(long long a,long long b){long long c=llgcd(a,b);a/=c;return a*b;}
long long llnCr(long long a,long long b){long long i,r=1;for(i=1;i<=b;i++){r*=(a+1-i);r/=i;}return r;}
long long llnHr(long long a,long long b){return llnCr(a+b-1,b);}
long long llfact(long long a){long long i,r=1;for(i=1;i<=a;i++){r*=i;}return r;}
long long llpow(long long a,long long b){long long i,r=1;for(i=1;i<=b;i++){r*=a;}return r;}
long long lldsum(long long x){long long r=0;while(x){r+=(x%10);x/=10;}return r;}
long long lldsumb(long long x,long long b){long long r=0;while(x){r+=(x%b);x/=b;}return r;}
long long llsankaku(long long x){return ((1+x)*x)/2;}

//double
double dbmax(double a,double b){if(a>b){return a;}return b;}
double dbmin(double a,double b){if(a<b){return a;}return b;}
double dbzt(double a,double b){return dbmax(a,b)-dbmin(a,b);}


typedef pair<int, int> ii;
typedef long long ll;
typedef vector<ll> vll;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<pll> vpll;
const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};

 
/*CODE START HERE*/
int main() {
  int h,w;
  cin >> h >> w;
  string s(h);
  for(int i= 0 ; i<h ; i++){
    cin >> s.at(i);
  }
  for (int i=0 ; i<h;i++){
    for(int j=0 ; j<w ; j++){
      if ( s.at(i).at(j)=='#') continue ;
      int  num = 0;
      for(int k=0; k<8;k++){
        int nx = (j+dx.at(k));
        int ny = (i+dy.at(k));
        
        if ( d.at(ny).at(nx) == '#'){
          num ++ ;
        }
        
      }
      s.at(ny).at(nx)= char(num + '0') ;
    }
  }
  for(int i=0 ;i<h ; i++){
    cout<< s(i) << endl;
  }
}
