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 MAXN = 1005;

int H, W, d[MAXN][MAXN];
char A[MAXN][MAXN];

int di[] = {1, -1, 0, 0};
int dj[] = {0, 0, 1, -1};

int main(){	
	scanf("%d %d", &H, &W);
	
	memset(d, -1, sizeof d);
	
	queue < pair <int, int> > q;
	for(int i = 0; i < H; i++){
		scanf("%s", A[i]);
		for(int j = 0; j < W; j++){
			if(A[i][j] == '#'){
				q.push(make_pair(i, j));
				d[i][j] = 0;
			}
		}
	}
	
	int mx = 0;
	
	while(!q.empty()){
		pair <int, int> u = q.front();
		q.pop();
		
		mx = max(mx, d[u.first][u.second]);
		
		for(int dir = 0; dir < 4; dir++){
			int ui = u.first + di[dir], uj = u.second + dj[dir];
			
			if(ui < 0 || ui >= H || uj < 0 || uj >= W)continue;
			if(d[ui][uj] != -1)continue;
			
			q.push(make_pair(ui, uj));
			d[ui][uj] = d[u.first][u.second] + 1;
		}
	}
	
	printf("%d\n", mx);
	
	return 0;
}
