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;
const int mx = 1005;
const int inf = 3000;
char a[mx][mx];
int n, m, up[mx][mx], ans = 0;

int main(void){
    
    cin>>n>>m;
    for(int i=1; i<=n; i++) scanf("%s", a[i]+1);
    for(int i=0; i<1003; i++) up[0][i] = up[i][0] = up[n+1][i] = up[i][m+1] = inf;
    for(int i=1; i<=n; i++) for(int j = 1; j <= m; j++)
        up[i][j] = a[i][j] == '#' ? 0 : min(up[i-1][j], up[i][j-1]) + 1;
    for(int i=n; i>0; i--) for(int j = m; j > 0; j--)
        up[i][j] = min(up[i][j], min(up[i+1][j], up[i][j+1]) + 1);
    for(int i=1; i<=n; i++) for(int j = 1; j <= m; j++) ans = max(ans, up[i][j]);
    cout<< ans <<endl;
    
}