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;
const int maxn = 1010;
int vis[maxn][maxn];
int main(){
	int n, m; cin >> n >> m;
	vector<string> g(n); for(auto & i : g) cin >> i;
	vector<pair<int,int> > q, nq;
	for(int i = 0; i < n; i++) for(int j = 0; j < m; j++) if(g[i][j] == '#') q.push_back({i, j}), vis[i][j] = 1;
	for(int i = 0; ; i++){
		for(auto e : q){
			int x, y; tie(x, y) = e;
			for(int di = -1; di <= 1; di++){
				for(int dj = -1; dj <= 1; dj++){
					if(abs(di) + abs(dj) != 1) continue;
					int nx = x + di, ny = y + dj;
					if(nx >= 0 && nx < n && ny >= 0 && ny < m && !vis[nx][ny]) nq.push_back({nx, ny}), vis[nx][ny] = 1;
				}
			}
		}
		if(nq.empty()){
			cout << i << endl;
			return 0;
		}
		q = nq;
		nq.clear();
	}
	return 0;
}
