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;
 
#define f(i,n) for (int i = 0; i < n; i++)
using pii = pair<int,int>;
 
int di[] = {0,0,1,-1};
int dj[] = {1,-1,0,0};
 
int main() {
	int n,m;
	cin >> n >> m;
	int dp[n][m];
	queue<pii> pq;
	f(i,n) {
		string s; cin >> s;
		f(j,m) {
			dp[i][j] = s[j] == '#'?0:n+m+1;
			if (!dp[i][j]) pq.push(pii(i,j));
		}
	}
	auto IN = [&](int i, int j) {
		return i>=0&&i<n&&j>=0&&j<m;
	};
	int ans = 0;
	while (pq.size()) {
		pii pp = pq.front(); pq.pop();
		int i = pp.first, j = pp.second;	
		f(k,4) {
			int ni = i+di[k], nj = j+dj[k];
			if (IN(ni,nj) && dp[ni][nj] == n+m+1) {
				dp[ni][nj] = dp[i][j]+1;
				pq.push(pii(ni,nj));
				ans = max(ans,dp[ni][nj]);
			}
		}
	}
	cout << ans << endl;
	return 0;
}
