题干：

求每个*能够到达的格子数量，只有.可以走（四个方向扩展），结果mod 10，替换 * 后输出。

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).

解题报告：

   直接对每个'.'进行搜索看连通块，然后对每一个'*'直接求答案就行了，注意不要算重复，所以需要set去判重。（因为有可能四个方向属于同一个联通块，所以不判重就会加四次）

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
const int MAX = 1000 + 5;
char maze[MAX][MAX];
int ans[MAX][MAX];
bool vis[MAX][MAX];
int f[MAX*MAX];
int num[MAX*MAX]; 
int n,m;
int nx[4] = {0,1,0,-1};
int ny[4] = {1,0,-1,0};
int getf(int v) {return v == f[v] ? v : f[v] = getf(f[v]);
}
void merge(int u,int v) {int t1 = getf(u);int t2 = getf(v);f[t2] = t1;
}
int get(int x,int y) {return (x-1)*m +y;
}
void go(int x,int y) {for(int k = 0; k<4; k++) {int tx = x + nx[k];int ty = y + ny[k];if(tx < 1 || tx > n || ty < 1 || ty > m) continue;if(maze[tx][ty] == '*' || vis[tx][ty]) continue;merge(get(x,y),get(tx,ty));vis[tx][ty] = 1;go(tx,ty);}
}
int main()
{cin>>n>>m;for(int i = 0; i<=n*m; i++) {f[i] = i;}for(int i = 1; i<=n; i++) {scanf("%s",maze[i]+1);}for(int i = 1; i<=n; i++) {for(int j = 1; j<=m; j++) {if(maze[i][j] == '.' && vis[i][j] == 0) {vis[i][j] = 1;//别忘这步！！虽然这题不会WA但是不写是错的。go(i,j);}			}}for(int i = 1; i<=n; i++) {for(int j = 1; j<=m; j++) {if(maze[i][j] == '*') continue;num[getf(get(i,j))]++;}}for(int i = 1; i<=n; i++) {for(int j = 1; j<=m; j++) {if(maze[i][j] == '.') ans[i][j] = '.';else {int cnt = 0;set<int> ss;for(int k = 0; k<4; k++) {int tx = i + nx[k];int ty = j + ny[k];if(maze[tx][ty] == '*')continue;if(tx < 1 || tx > n || ty < 1 || ty > m) continue;ss.insert(getf(get(tx,ty)));}auto it = ss.begin();for(;it!=ss.end();++it) cnt += num[*it];ans[i][j] = cnt;}}}for(int i = 1; i<=n; i++) {for(int j = 1; j<=m; j++) {if(maze[i][j] == '.') printf(".");else printf("%d",(ans[i][j]+1)%10);}puts("");}return 0 ;
}