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;

int N, M;
char arr[1004][1004];
int q[1004*1004][4], f, r;
int wx[5] = {-1, 0, 1, 0}, wy[5] = {0, 1, 0, -1};

int main(){
    scanf("%d %d", &N, &M);
    for(int i = 1; i <= N; ++i) scanf("%s", arr[i]+1);
    for(int i = 1; i <= N; ++i) for(int j = 1; j <= M; ++j) if(arr[i][j] == '#'){
        q[r][0] = i, q[r][1] = j, q[r++][2] = 0;
    }
    while(f < r){
        for(int i = 0; i < 4; ++i){
            int nx = q[f][0] + wx[i], ny = q[f][1] + wy[i];
            if(arr[nx][ny] == '.'){
                q[r][0] = nx, q[r][1] = ny, q[r++][2] = q[f][2] + 1;
                arr[nx][ny] = '#';
            }
        }
        ++f;
    }
    printf("%d\n", !f ? 0 : q[f-1][2]);
    return 0;
}