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 main(){
	int H,W;
	cin >> H >> W;
	vector<string> A(H);
	int m[H][W];
	for(int i=0;i<H;i++)cin >> A[i];
	queue<int> que;
	for(int i=0;i<H;i++)for(int j=0;j<W;j++){
		if(A[i][j]=='#'){
			que.push(i*W+j);
			m[i][j]=0;
		}else m[i][j]=9999999;
	}
	
	int dx[4]={0,1,0,-1};
	int dy[4]={1,0,-1,0};
	while(!que.empty()){
		int a=que.front();
		que.pop();
		int h=a/W;
		int w=a%W;
		for(int i=0;i<4;i++){
			int hh=h+dx[i];
			int ww=w+dy[i];
			if(0<=hh && hh<H && 0<=ww && ww<W && (m[hh][ww]>m[h][w]+1)){
				m[hh][ww]=m[h][w]+1;
				que.push(hh*W+ww);
			}
		}
	}
	int maxi=0;
	for(int i=0;i<H;i++)for(int j=0;j<W;j++)maxi=max(maxi,m[i][j]);
	cout << maxi << endl;
	return 0;
}