You are given a grid of squares with H horizontal rows and W vertical columns, where each square is painted white or black. HW characters from A_{11} to A_{HW} represent the colors of the squares. A_{ij} is `#` if the square at the i-th row from the top and the j-th column from the left is black, and A_{ij} is `.` if that square is white.

We will repeatedly perform the following operation until all the squares are black:

* Every white square that shares a side with a black square, becomes black.



Find the number of operations that will be performed. The initial grid has at least one black square.

Constraints

* 1 \leq H,W \leq 1000
* A_{ij} is `#` or `.`.
* The given grid has at least one black square.

Input

Input is given from Standard Input in the following format:


H W
A_{11}A_{12}...A_{1W}
:
A_{H1}A_{H2}...A_{HW}


Output

Print the number of operations that will be performed.

Examples

Input

3 3
...
.#.
...


Output

2


Input

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


Output

3
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef double D;
typedef pair<int,int> P;
#define M 1000000007
#define F first
#define S second
#define PB push_back
#define INF 100000000000000000
int h,w,d[1005][1005],dx[4]={0,1,0,-1},dy[4]={1,0,-1,0};
int main(void){
    scanf("%d%d",&h,&w);
    queue<P>bfs;
    for(int i=0;i<h;i++){
		for(int j=0;j<w;j++){
			char c;
			scanf(" %c",&c);
			if(c=='#')bfs.push(P(i,j));
			else d[i][j]=-1;
		}
	}
	while(!bfs.empty()){
		int y=bfs.front().F,x=bfs.front().S;
		bfs.pop();
		for(int k=0;k<4;k++){
			int ny=y+dy[k],nx=x+dx[k];
			if(0<=ny&&ny<h&&0<=nx&&nx<w&&d[ny][nx]==-1){
				d[ny][nx]=d[y][x]+1;
				bfs.push(P(ny,nx));
			}
		}
	}
	int ans=0;
	for(int i=0;i<h;i++)for(int j=0;j<w;j++)ans=max(ans,d[i][j]);
	printf("%d\n",ans);
}
