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;
int g[1000][1000];
string s[1000];
int n,m;
inline bool in(int x,int y){
	return x>=0&&x<n&&y>=0&&y<m;
}
int dx[4] = {1,0,-1,0};
int dy[4] = {0,-1,0,1};
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cin>>n>>m;
	memset(g,-1,sizeof g);
	for(int i=0;i<n;i++)
		cin>>s[i];
	queue<pair<int,int> > q;
	for(int i=0;i<n;i++)
		for(int j=0;j<m;j++)
			if(s[i][j]=='#')
				g[i][j]=0,q.emplace(i,j);
	while(!q.empty()){
		int x,y; tie(x,y) = q.front();q.pop();
		for(int i=0;i<4;i++){
			int nx=x+dx[i],ny=y+dy[i];
			if(in(nx,ny)&&g[nx][ny]==-1){
				g[nx][ny]=g[x][y]+1;
				q.emplace(nx,ny);
			}
		}
	}
	int ans=0;
	for(int i=0;i<n;i++)
		for(int j=0;j<m;j++)
			ans=max(ans,g[i][j]);
	cout<<ans<<endl;
		
}
