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>
#define ll long long
using namespace std;
char a[1050][1050],c;
int mx[4] = {0,1,0,-1},my[4]={1,0,-1,0};
struct node{
	int x,y,st;
	node(int _x,int _y,int _z){
		x = _x;y = _y;st = _z;
	}
	node(){
	};
};
queue<node> q;
ll h,w;
int main(){
	ios::sync_with_stdio(false);
	cin >> h >> w;
	for(int i = 1;i <= h;i ++){
		for(int j = 1;j <= w;j ++){
			cin >>c;
			a[i][j] = c;
			if(a[i][j] == '#') q.push(node(i,j,1));
		}
	}
	node p;
	while(!q.empty()){
		p = q.front();
		q.pop();
		for(int i=0;i < 4;i ++)
			if(a[p.x+mx[i]][p.y+my[i]] == '.'){
				q.push(node(p.x+mx[i],p.y+my[i],p.st+1));
				a[p.x+mx[i]][p.y+my[i]] = '#';
			}		
	}

	cout << p.st-1 << endl;
    return 0;
}
