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;

int main()
{
	int h, w;
	cin >> h >> w;
	vector<string> a(h);
	for(auto&& i : a)
	{
		cin >> i;
	}
	queue<tuple<int, int, int>> que;
	for(int i = 0; i < h; i++)
	{
		for(int j = 0; j < w; j++)
		{
			if(a[i][j] == '#')
			{
				que.push(make_tuple(i, j, 0));
			}
		}
	}
	int ans = 0;
	int dx[] = { 1, 0, -1, 0 };
	int dy[] = { 0, 1, 0, -1 };
	while(!que.empty())
	{
		auto status = que.front();
		que.pop();
		
		int x = get<0>(status);
		int y = get<1>(status);
		ans = get<2>(status);
		for(int i = 0; i < 4; i++)
		{
			if (x + dx[i] < 0 || y + dy[i] < 0 || x + dx[i] == h || y + dy[i] == w) continue;
			if (a[x + dx[i]][y + dy[i]] == '#') continue;
			a[x + dx[i]][y + dy[i]] = '#';
			que.push(make_tuple(x + dx[i], y + dy[i], ans + 1));
		}
	}
	cout << ans << "\n";
}
