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<bits/stdc++.h>

using namespace std;

signed main(){
	int h, w; cin>>h>>w;
	vector<vector<int> > a(h, vector<int>(w));
	vector<vector<int> > vis(h, vector<int>(w, -1));
	queue<pair<int, int> > q;
	for(int i = 0; i < h; i++){
		for(int j =0 ; j < w; j++){
			char c; cin>>c;
			if(c == '#'){
				a[i][j] = 1;
				vis[i][j] = 0;
				q.push({i, j});
			}else{
				a[i][j] = 0;
			}
		}
	}
	int count = 0;
	while(!q.empty()){
		int i = q.front().first, j = q.front().second;
		q.pop();
		vector<int> dx = {1, 0, -1, 0};
		vector<int> dy = {0, 1, 0, -1};
		for(int k = 0; k < 4; k++){
			int nx = dx[k] + i;
			int ny = dy[k] + j;
			if(!(nx >= 0 && nx < h && ny >= 0 && ny < w)) continue;
			if(vis[nx][ny] >= 0) continue;
			q.push({nx, ny});
			vis[nx][ny] = vis[i][j] + 1;
			count = max(count, vis[nx][ny]);
		}
	}
	cout<<count<<endl;
}