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 X first
#define Y second
using namespace std;
typedef pair<int,int> pii;

int h, w, d[1005][1005], ans;
char tmp[1005];

queue<pii> q;

bool valid (int X, int Y) {
	return (1 <= X && X <= h && 1 <= Y && Y <= w);
}

int main()
{
	scanf("%d%d",&h,&w);
	for(int i=1;i<=h;i++) {
		scanf("%s",tmp+1);
		for(int j=1;j<=w;j++) {
			if(tmp[j] == '#') {
				d[i][j] = 1;
				q.push({i, j});
			}
		}
	}
	while(!q.empty()) {
		int A, B;
		tie(A, B) = q.front();
		ans = max(ans, d[A][B]-1);
		q.pop();
		for(int i=0;i<4;i++) {
			int dx = "1210"[i] - '1', dy = "2101"[i] - '1';
			if(!valid(A+dx, B+dy) || d[A+dx][B+dy]) continue;
			d[A+dx][B+dy] = d[A][B] + 1;
			q.push({A+dx, B+dy});
		}
	}
	printf("%d\n",ans);
}
