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 R,C,Time=-1,Piv,tmpx,tmpy;
int dx[4]={0,1,0,-1};
int dy[4]={1,0,-1,0};
char grid[1005][1005];
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	cin>>R>>C;
	queue<pair<int,int> > bfs;
	for(int i=0;i<=R+1;++i){
		for(int j=0;j<=C+1;++j){
			if(i==0||i==R+1||j==0||j==C+1){
				grid[i][j]='X';
				continue;
			}
			cin>>grid[i][j];
			if(grid[i][j]=='#') bfs.push({i,j});
		}
	}
	while(!bfs.empty()){
		Piv=bfs.size();
		for(int i=0;i<Piv;++i){
			tmpx=bfs.front().first;
			tmpy=bfs.front().second;
			bfs.pop();
			for(int j=0;j<4;++j){
				if(grid[tmpx+dx[j]][tmpy+dy[j]]=='.'){
					grid[tmpx+dx[j]][tmpy+dy[j]]='#';
					bfs.push({tmpx+dx[j],tmpy+dy[j]});
				}
			}
		}
		Time++;
	}
	cout<<Time<<endl;
}