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 <iostream>
#include <queue>
using namespace std;

const int dx[4] = { 1, -1, 0, 0 }, dy[4] = { 0, 0, 1, -1 }, MAXN = 1005;
char mp[MAXN][MAXN];
queue<pair<pair<int, int>, int> > Q;
bool se[MAXN][MAXN];
int n, m, ans;

bool isval(int x, int y) {
	return x >= 0 && x < n && y >= 0 && y < m;
}
void addv(pair<int, int> v, int d) {
	for (int i = 0; i < 4; i++) {
		int nx = v.first + dx[i], ny = v.second + dy[i];
		if (isval(nx, ny) && !se[nx][ny])
			Q.push({ { nx, ny }, d + 1 }), se[nx][ny] = 1;
	}
}

int main() {
	cin >> n >> m;
	for (int i = 0; i < n; i++)
		for (int j = 0; j < m; j++) {
			cin >> mp[i][j];
			if (mp[i][j] == '#')
				Q.push({ {i, j}, 0 }), se[i][j] = 1;
		}
	while (!Q.empty()) {
		addv(Q.front().first, Q.front().second);
		ans = max(ans, Q.front().second);
		Q.pop();
	}
	cout << ans;
}