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;
const int maxn=1e3+1;

int vis[maxn][maxn],ans[maxn][maxn];
int res,H,W;
queue<pair<int,int>>q;
char S[maxn];

int main(){
	scanf("%d%d",&H,&W);
	for (int i=1;i<=H;++i){
		scanf("%s",S+1);
		for (int j=1;j<=W;++j)
		if (S[j]=='#') q.push(make_pair(i,j)),vis[i][j]=1;
	}
	while(!q.empty()){
		pair<int,int>u=q.front();
		q.pop();
		int x=u.first,y=u.second;
		res=max(res,ans[x][y]);
		if (x>1) if (!vis[x-1][y])
		vis[x-1][y]=1,q.push(make_pair(x-1,y)),
		ans[x-1][y]=ans[x][y]+1;
		if (y>1) if (!vis[x][y-1])
		vis[x][y-1]=1,q.push(make_pair(x,y-1)),
		ans[x][y-1]=ans[x][y]+1;
		if (x<H) if (!vis[x+1][y])
		vis[x+1][y]=1,q.push(make_pair(x+1,y)),
		ans[x+1][y]=ans[x][y]+1;
		if (y<W) if (!vis[x][y+1])
		vis[x][y+1]=1,q.push(make_pair(x,y+1)),
		ans[x][y+1]=ans[x][y]+1;
	}
	cout<<res<<endl;
}