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 Rint register int
using namespace std;
typedef pair<int, int> pii;
const int N = 1003, dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};
int n, m, front, rear, ans;
struct Node {
	int x, y, dep;
} q[N * N];
bool vis[N][N];
char str[N];
int main(){
	scanf("%d%d", &n, &m);
	for(Rint i = 1;i <= n;i ++){
		scanf("%s", str + 1);
		for(Rint j = 1;j <= m;j ++)
			if(str[j] == '#'){
				vis[i][j] = true;
				q[rear ++] = (Node){i, j, 0};
			}
	}
	while(front < rear){
		Node now = q[front ++];
		ans = max(now.dep, ans);
		for(Rint i = 0;i < 4;i ++){
			int nx = now.x + dx[i], ny = now.y + dy[i];
			if(nx >= 1 && nx <= n && ny >= 1 && ny <= m && !vis[nx][ny]){
				q[rear ++] = (Node){nx, ny, now.dep + 1};
				vis[nx][ny] = true;
			}
		}
	}
	printf("%d", ans);
}