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

const int INF = 1000000000;

int main() {
	int h, w; cin >> h >> w;
	vector<string>A(h); for (auto&& x : A)cin >> x;
	vector<vector<int>>G(h, vector<int>(w, INF));
	queue<pair<int, int>>que;

	for (int i = 0; i < h; i++) {
		for (int j = 0; j < w; j++) {
			if (A[i][j] == '#') {
				G[i][j] = 0;
				que.push({ i,j });
			}
		}
	}

	int ans = 0;
	int dx[4] = { -1,0,1,0 }, dy[4] = { 0,1,0,-1 };
	while (!que.empty()) {
		int x = que.front().second, y = que.front().first;
		que.pop();
		for (int i = 0; i < 4; i++) {
			int nx = x + dx[i], ny = y + dy[i];
			if (0 > nx || 0 > ny || nx >= w || ny >= h || G[ny][nx] != INF)continue;
			G[ny][nx] = G[y][x] + 1;
			que.push({ ny,nx });
			ans = max(ans, G[ny][nx]);
		}
	}
	cout << ans << endl;

	return 0;
}