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 <stdio.h>
#include <queue>
using namespace std;

int N, M, D[1010][1010]; char S[1010][1010];

int main()
{
	scanf ("%d %d", &N, &M);
	for (int i = 0; i < N; i++) scanf ("%s", S[i]);

	queue< pair<int, int > > Q;
	for (int i = 0; i < N; i++) for (int j = 0; j < M; j++) if (S[i][j] == '#') Q.push({ i,j });

	int a = 0;
	while (!Q.empty()){
		int x = Q.front().first, y = Q.front().second; Q.pop();
		int d = D[x][y]; a = d;
		int dx[4] = { 0,1,0,-1 };
		int dy[4] = { 1,0,-1,0 };
		for (int k = 0; k < 4; k++){
			int px = x + dx[k];
			int py = y + dy[k];
			if (px < 0 || px >= N || py < 0 || py >= M) continue;
			if (S[px][py] == '.'){
				Q.push({ px,py }); S[px][py] = '#'; D[px][py] = d + 1;
			}
		}
	}

	printf ("%d\n", a);
	return 0;
}
