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
/*
オーダー
10**6	余裕を持って間に合う	
10**7	おそらく間に合う	余裕を持って間に合う
10**8	非常にシンプルな処理でない限り厳しい	おそらく間に合う
10**9		非常にシンプルな処理でない限り厳しい
logn	:OK
n		:10^7
nlogn	:10^6
n**2	:10^4
n**3	:300
2**n	:20
n!		:10



// 各桁の和を計算する関数
int findSumOfDigits(int n) {
  int sum = 0;
  while (n > 0) { // n が 0 になるまで
    sum += n % 10;
    n /= 10;
  }
  return sum;
}

sort(a, a + N, greater<int>());  // a[0:N] を大きい順にソート

int num[110] = {0};  // バケット
for (int i = 0; i < N; ++i) {
    num[d[i]]++;  // d[i] が 1 個増える
}

map<string, int> mp; // 連想配列 map<キー型, 値型> オブジェクト名
for (int i = 0; i < N; ++i) {
    auto itr = mp.find(s[i]);        // s[i] が設定されているか？
    if(itr != mp.end() ) {
         mp[s[i]] += 1;
    }
    else {
         mp[s[i]] += 1 ;
    }
}
stack<int> s;	//intをデータとするスタックを用意
s.push(1);		//{} -> {1}
printf("%d\n", s.top());	// 3
s.pop();

queue<int> que;	//intをデータとするキューを用意
que.push(1);		//{} -> {1}
printf("%d\n", que.front());	// 1
que.pop();
*/

#include <iostream>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <bitset>
#include <string.h>

using namespace std;

// #define for(i,a,b) for (int i=(a);i<(b);++i)
typedef long long ll;
typedef pair<ll,ll> P;
#define REP(i,n) for(long long i = 0; i < (long long)(n); i++)
#define pb push_back // vectorに要素追加
#define INF (ll)1e18
// int 
// // 各桁の和を計算する関数
// int findSumOfDigits(int n) {
//   int amari = 0;
//   int keta = 0;
//   while (n > 0) { // n が 0 になるまで
//     amari += n % 2;
//     if (keta%2==0)
//     n /= 10;
//   }
//   return sum;
// }
 
ll gcd(ll a, ll b) {
	if(b == 0) return a;
	return gcd(b, a % b);
}

// const ll MAX_N = 1e5;
// ll N;
// ll h[MAX_N];
// ll dp[MAX_N+1]; // メモ化テーブル
 
// // i番目以降の品物から重さの総和がj以下になるように選ぶ
// void rec(ll i){
// 	if (i==0) dp[i] = 0;
// 	else if (i==1) dp[i] = abs(h[1]-h[0]);
// 	else dp[i] = min(dp[i-1]+abs(h[i]-h[i-1]), dp[i-2]+abs(h[i]-h[i-2]));
// 	return;
// }
 
 
int main() {
	// 入力
	ll H,W;
	cin >>H>>W;
	string S[H];
	ll map[H][W] = {0};
	REP(i,H)REP(j,W) map[i][j] = 0;
	for (ll i = 0; i < H; ++i) cin >> S[i];
 
	// 解法
	REP(i,H){
		REP(j,W){
			if (S[i][j]=='#'){
				map[i-1][j-1]++;map[i][j-1]++;map[i+1][j-1]++;
				map[i-1][j  ]++;map[i][j  ]++;map[i+1][j  ]++;
				map[i-1][j+1]++;map[i][j+1]++;map[i+1][j+1]++;
			}
		}
	}
	
	REP(i,H){
		REP(j,W){
			if (S[i][j]=='.') printf("%lld", map[i][j]);
			else  printf("#");
		}
		printf("\n");
	}
}