Problem: 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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;

template <typename T> void chmax(T &x,const T &y)
{
	if(x<y)x=y;
}
#define rep(i,l,r) for(int i=l;i<=r;++i)
const int N=1000+5,dx[]={-1,1,0,0},dy[]={0,0,-1,1};
char s[N][N];
int g[N][N];

int main()
{
#ifdef kcz
	freopen("1.in","r",stdin);
#endif
	int n,m;
	cin>>n>>m;
	rep(i,1,n)scanf("%s",s[i]+1);
	
	deque<pair<int,int>>q;
	rep(i,1,n)
	rep(j,1,m)
	if(s[i][j]=='#')q.push_back({i,j});
	else g[i][j]=-1;
	
	int ans=0;
	while(!q.empty())
	{
		int x,y;
		tie(x,y)=q.front();q.pop_front();
		chmax(ans,g[x][y]);
		rep(i,0,3)
		if(g[x+dx[i]][y+dy[i]]==-1)
		{
			g[x+dx[i]][y+dy[i]]=g[x][y]+1;
			q.push_back({x+dx[i],y+dy[i]});
		}
	}
	
	cout<<ans;
}