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 <iostream>
#include <queue>
#include <string.h>
using namespace std;
const int MAXN=1005;
const int dx[]={-1,1,0,0};
const int dy[]={0,0,-1,1};
int n,m;
bool a[MAXN][MAXN];
int vis[MAXN][MAXN];

int main()

{
	memset(vis,-1,sizeof(vis));
	queue<pair<int,int> > q;
	cin>>n>>m;
	for (int i=1;i<=n;i++){
		for (int j=1;j<=m;j++){
			char c;
			cin>>c;
			a[i][j]=c=='#';
			if (a[i][j]){
				q.push(make_pair(i,j));
				vis[i][j]=0;
			}
		}
	}

	while (!q.empty()){
		int x=q.front().first,y=q.front().second;
		q.pop();
		for (int i=0;i<4;i++){
			int gx=x+dx[i],gy=y+dy[i];
			if (gx<=0 || gy<=0 || gx>n || gy>m) continue;
			if (vis[gx][gy]==-1){
				vis[gx][gy]=vis[x][y]+1;
				q.push(make_pair(gx,gy));
			}
		}
	}

	int ans=0;
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++)
			ans=max(ans,vis[i][j]);
			
	cout<<ans<<endl;
	return 0;
}