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;

#define PII pair < int , int >
const int dir[4][2] = {0,1,0,-1,1,0,-1,0};
queue < PII > q; int N , M , dist[1003][1003];

int main(){
	cin >> N >> M; memset(dist , -1 , sizeof(dist));
	for(int i = 1 ; i <= N ; ++i)
		for(int j = 1 ; j <= M ; ++j){
			char c; cin >> c;
			if(c == '#'){dist[i][j] = 0; q.push(PII(i , j));}
		}
	int mx = 0;
	while(!q.empty()){
		PII t = q.front(); q.pop();
		for(int i = 0 ; i < 4 ; ++i){
			int x = t.first + dir[i][0] , y = t.second + dir[i][1];
			if(x && x <= N && y && y <= M && dist[x][y] == -1){
				mx = max(mx , dist[x][y] = dist[t.first][t.second] + 1);
				q.push(PII(x , y));
			}
		}
	}
	cout << mx; return 0;
}