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;

const int N = 1005, dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1};

int n, m, d[N][N], r;
char b[N][N];
queue<tuple<int, int>> q;

int main(){
	scanf("%d%d", &n, &m);
	memset(d, 0x7f, sizeof(d));
	for(int i = 1; i <= n; i++){
		scanf("%s", b[i] + 1);
		for(int j = 1; j <= m; j++){
			if(b[i][j] == '#'){
				d[i][j] = 0;
				q.emplace(i, j);
			}
		}
	}
	while(!q.empty()){
		int x, y; tie(x, y) = q.front(); q.pop();
		r = max(r, d[x][y]);
		for(int i = 0; i < 4; i++){
			int nx = x + dx[i], ny = y + dy[i];
			if(!b[nx][ny]) continue;
			if(d[nx][ny] > d[x][y] + 1){
				d[nx][ny] = d[x][y] + 1;
				q.emplace(nx, ny);
			}
		}
	}
	printf("%d\n", r);
}