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<stdio.h>
#include<string.h>
#define N 1005
char A[N][N];
int f[N*N][3],res=0;
int main() {
	int n,m,i,j,h=0,t=0,ans,x,y;
	int fs[8]={-1,0,0,-1,0,1,1,0}; 
	scanf("%d%d",&n,&m);
	for(i=0;i<n;i++) {
		getchar();
		for(j=0;j<m;j++) {
			scanf("%c",&A[i][j]);
			if(A[i][j]=='#') {
				f[t][0]=i,
				f[t][1]=j,
				f[t][2]=0;
				t++;
			}
			else res++;
		}
	}
	while(h<t) {
		for(i=0;i<8;i+=2) {
			x=fs[i]+f[h][0],
			y=fs[i+1]+f[h][1];
			if(x>=0 && y>=0 && x<n && y<m && A[x][y]=='.') {
				A[x][y]='#';
				res--;
				f[t][0]=x,f[t][1]=y,f[t][2]=f[h][2]+1;
				t++;
			}
		}
		h++;
		if(!res) break;
	}
	printf("%d",f[t-1][2]);
} 