Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define PII pair <int, int>
#define x first
#define y second

using namespace std;

int n, m;

char ch[1010][1010];

vector <PII> V(0);

int Dis[1010][1010];

int main() {
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; i++) scanf("%s", ch[i] + 1);
	memset(Dis, -1, sizeof Dis);
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= m; j++)
			if(ch[i][j] == '#') V.push_back(PII(i, j)), Dis[i][j] = 0;
	int ans = 0;
	for(int i = 0; i < V.size(); i++) {
		PII S = V[i];
		for(int j = -1; j <= 1; j++)
			for(int k = -!j; k <= 1; k += 2) {
				PII K = PII(S.x + j, S.y + k);
				if(K.x >= 1 && K.x <= n && K.y >= 1 && K.y <= m && Dis[K.x][K.y] == -1) {
					ans = (Dis[K.x][K.y] = Dis[S.x][S.y] + 1);
					V.push_back(K);
				}
			}
	}
	printf("%d\n", ans);
	return 0;
}