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;

void solve(void)
{
	int h, w;
	cin >> h >> w;
	vector<string> grid(h);
	for (int i = 0; i < h; i++)
	{
		cin >> grid[i];
	}
	vector<int> dis(h * w, -1);
	queue<int> q;
	for (int i = 0; i < h; i++)
		for (int j = 0; j < w; j++)
		{
			if (grid[i][j] == '#')
			{
				dis[i*w+j] = 0;
				q.push(i * w + j);
			}
		}
	int d[] = {0, 1, 0, -1};
	while (!q.empty())
	{
		int y = q.front() / w, x = q.front() % w;
		q.pop();
		for (int i = 0; i < 4; i++)
		{
			int ny = y + d[i], nx = x + d[i ^ 1];
			if (0 <= ny && ny < h && 0 <= nx && nx < w)
			{
				if (dis[ny*w+nx] == -1)
				{
					dis[ny * w + nx] = dis[y * w + x] + 1;
					q.push(ny * w + nx);
				}
			}
		}
	}
	cout << *max_element(dis.begin(), dis.end()) << endl;
}

int main()
{
	solve();
	//cout << "yui(*-v・)yui" << endl;
	return 0;
}
