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;

int H, W, dist[1010][1010], final_res;
char grid[1010][1010];
queue<pair<int, int > > Q;

int mv1[4] = {-1, 1, 0, 0};
int mv2[4] = {0, 0, 1, -1};

int main(){
	//freopen("test.in", "r+", stdin);

	cin >> H >> W;
	for(int i = 0; i < H; i++){
		for(int j = 0; j < W; j++){
			cin >> grid[i][j];
			dist[i][j] = 1e9;
			if(grid[i][j] == '#'){
				Q.push(make_pair(i, j));
				dist[i][j] = 0;
			}
		}
	}
	while(!Q.empty()){
		int i = Q.front().first;
		int j = Q.front().second;
		Q.pop();

		for(int k = 0; k < 4; k++){
			int ii = i + mv1[k];
			int jj = j + mv2[k];
			if(dist[ii][jj] == 1e9){
				Q.push(make_pair(ii, jj));
				dist[ii][jj] = dist[i][j] + 1;
				final_res = max(final_res, dist[ii][jj]);
			}
		}
	}
	cout << final_res << endl;
}