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 <cstdio>
#include <queue>
#include <utility>
typedef std::pair<int,int> P;
typedef std::pair<int,P> node;
const int way[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
char s[1005][1005];
int main() {
	int h,w;scanf("%d%d",&h,&w);
	int ans=0;std::queue<node> q;
	for(register int i=1;i<=h;++i) {
		scanf("%s",s[i]+1);
		for(register int j=1;j<=w;++j) {
			if(s[i][j]=='#') q.push(node(0,P(i,j)));
		}
	}
	while(q.size()) {
		node cur=q.front();q.pop();
		for(register int i=0;i<4;++i) {
			int x=cur.second.first+way[i][0];
			int y=cur.second.second+way[i][1];
			if(1<=x&&x<=h&&1<=y&&y<=w&&s[x][y]=='.') {
				s[x][y]='#';q.push(node(cur.first+1,P(x,y)));
			}
		}
		ans=std::max(ans,cur.first);
	}
	printf("%d\n",ans);
	return 0;
}
