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
//ihdignite

#include <bits/stdc++.h>

using namespace std;

typedef pair<int, int> pii;

const int MAXN = 1001, di[4]={1, 0, -1, 0}, dj[4]={0, 1, 0, -1};

int h, w, d[MAXN][MAXN], ans;
string a[MAXN];
queue<pii> qu;

int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(NULL);

	cin >> h >> w;
	memset(d, 0x3f, sizeof(d));
	for(int i = 0; i < h; i++) {
		cin >> a[i];
		for(int j = 0; j < w; j++) {
			if(a[i][j] == '#') {
				qu.push(pii(i, j));
				d[i][j]=0;
			}
		}
	}

	while(!qu.empty()) {
		pii u = qu.front();
		qu.pop();
		ans = max(ans, d[u.first][u.second]);
		for(int k = 0; k < 4; ++k) {
			int ni = u.first+di[k], nj = u.second+dj[k];
			if(ni>=0&&ni<h&&nj>=0&&nj<w&&d[ni][nj]>1e9) {
				d[ni][nj] = d[u.first][u.second]+1;
				qu.push(pii(ni, nj));
			}
		}
	}

	cout << ans << '\n';

	return 0;
}
