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 dr[] = {-1, 0, 1, 0}, dc[] = {0, 1, 0, -1};

int h, w;
int ans;
char grid[1024][1024];
int dist[1024][1024];

int main() {
	queue<pair<int, int>> q;
	cin >> h >> w;
	for (int i = 0; i < h; i++)
	for (int j = 0; j < w; j++)
	{
		cin >> grid[i][j];
		if (grid[i][j] == '#') 
			q.emplace(i, j);
	}
	
	while (q.size()) {
		int r, c;
		tie(r, c) = q.front();
		q.pop();
		for (int i = 0; i < 4; i++) {
			int nr, nc;
			nr = r + dr[i];
			nc = c + dc[i];
			if (nr < 0 || nr >= h || nc < 0 || nc >= w) continue;
			if (grid[nr][nc] == '#') continue;
			grid[nr][nc] = '#';
			dist[nr][nc] = dist[r][c] + 1;
			ans = max(ans, dist[nr][nc]);
			q.emplace(nr, nc);
		}
	}
	cout << ans << '\n';
	return 0;
}
