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;
#define pi pair<int,int>
const int N=1e3+10;
int vis[N][N];
char A[N][N];
int dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
queue<pi>Q;
int main()
{
	int n,m;scanf("%d%d",&n,&m);
	for(int i=0;i<n;i++){
		scanf(" %s",A[i]);
		for(int j=0;j<m;j++)if(A[i][j]=='#')vis[i][j]=1,Q.push({i,j});
	}
	int mx=1;
	while(!Q.empty()){
		auto p=Q.front();Q.pop();
		//cout<<p.first<<' '<<p.second<<'\n';
		for(int i=0;i<4;i++){
			int x=p.first+dir[i][0],y=p.second+dir[i][1];
			if(x<0||y<0||x>=n||y>=m||vis[x][y])continue;
			vis[x][y]=vis[p.first][p.second]+1;
			Q.push({x,y});
			mx=max(vis[x][y],mx);
		}
	}cout<<mx-1<<'\n';
}
