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 <iostream>
#include <vector>
#include <algorithm>
#include <queue>
using namespace std;
int H,W;
char G[1001][1001];
int ans=0;
int dh[4]={0,-1,0,1};
int dw[4]={1,0,-1,0};
queue< pair < pair<int,int> , int > > que;
int main(){
	cin >> H >> W;
	for(int i=0;i<H;i++)cin >> G[i];
	for(int i=0;i<H;i++){
		for(int j=0;j<W;j++){
			if(G[i][j] == '#') que.push(make_pair(make_pair(i,j),0));
		}
	}
	
	while(!que.empty()){
		int h,w,t;
		h= que.front().first.first;
		w= que.front().first.second;
		t= que.front().second;
		que.pop();
		for(int i=0;i<4;i++){
			int hh=h+dh[i];
			int ww=w+dw[i];
			if(0<=hh && hh<H && 0<=ww & ww<W && G[hh][ww]=='.'){
				G[hh][ww]='#';
				que.push(make_pair(make_pair(hh,ww),t+1));
				ans=t+1;
			}
		}
	}
	cout << ans << endl;
	return 0;
}