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 n , m , res;
int dx[4] = {-1,+1,0,0};
int dy[4] = {0,0,-1,+1};
int op[1005][1005];
char c[1005][1005];
queue<pair<int,int > > q; 
bool check(int x, int y){
	return (x >= 1 && x <= n && y >= 1 && y <= m);
}
int main(){
	cin >> n >> m;
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= m; j ++){
			cin >> c[i][j];
			op[i][j] = -1;
			if(c[i][j] == '#'){
				q.push({i,j}), op[i][j] = 0;
			}
		}
	}
	while(q.size()){
		int x = q.front().first;
		int y = q.front().second;
		q.pop();
		for(int i = 0; i < 4; i ++)
			if(op[x + dx[i]][y + dy[i]] == -1 && (check(x + dx[i] , y + dy[i]))){
				q.push({x + dx[i],y + dy[i]});
				op[x + dx[i]][y + dy[i]] = op[x][y] + 1; 
			}
	} 
	for(int i = 1; i <= n; i ++){
		for(int j = 1; j <= m; j ++){
			res = max(res , op[i][j]);
		}
	}
	cout << res << "\n";
}