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;
int main(){
	int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};
	int h,w;
	cin>>h>>w;
	vector<string>a(h);
	vector<pair<int,int>>newblack,nextblack;
	for(int i=0;i<h;i++){
		cin>>a[i];
		for(int j=0;j<w;j++)
			if(a[i][j]=='#')
				nextblack.emplace_back(i,j);
	}
	bool update;
	for(int ans=0;;ans++){
		update=false;
		newblack=nextblack;
		nextblack.clear();
		for(auto e:newblack){
			for(int i=0;i<4;i++){
				int nx=e.first+dx[i],ny=e.second+dy[i];
				if(0<=nx&&nx<h&&0<=ny&&ny<w&&a[nx][ny]=='.'){
					update=true;
					a[nx][ny]='#';
					nextblack.emplace_back(nx,ny);
				}
			}
		}
		if(update==false){
			cout<<ans<<endl;
			return 0;
		}
	}
}