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;

int main(){
	int n,m;
	cin >> n >> m;
	vector<vector<bool>> seen(n+2,vector<bool>(m+2,true));
	vector<pair<int,int>> all = {{1,0},{0,1},{-1,0},{0,-1}};
	char ch[n+2][m+2];
	queue<pair<int,int>> q;
	for(int i = 1 ; i <= n ; i += 1){
		for(int j = 1 ; j <= m ; j += 1){
			char a;
			cin >> a;
			ch[i][j] = a;
			seen[i][j] = false;
			if(a == '#'){
				q.push({i,j});
				seen[i][j] = true;
			}
		}
	}
	int ans = -1;
	while(q.size() > 0){
		ans += 1;
		int t = q.size();
		while(t--){
			pair<int,int> val = q.front();
			q.pop();
			seen[val.first][val.second] = true;
			for(pair<int,int> i : all){
				int v1 = val.first + i.first;
				int v2 = val.second + i.second;
				if(!seen[v1][v2] && ch[v1][v2] == '.'){
					q.push({v1,v2});
					seen[v1][v2] = true;
				}
			}
		}
	}
	cout << ans << endl;
}