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>
#define NMAX 1001

using namespace std;
struct pnt{int x,y,d;};
queue<pnt> q;
int dl[]={0,1,0,-1},dc[]={1,0,-1,0};
char b[NMAX][NMAX];
int dmax=0,n,m;
void bfs(){
    while(q.size()){
        dmax=max(dmax,q.front().d);
        for(int i=0;i<4;i++){
            int X=q.front().x+dc[i],Y=q.front().y+dl[i];
            if(X>=0 && X<n && Y>=0 && Y<m & b[X][Y]=='.'){
                b[X][Y]='#';
                q.push({X,Y,q.front().d+1});
            }
        }
        q.pop();
    }
}
int main(){
    ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
    cin>>n>>m;
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            cin>>b[i][j];
            if(b[i][j]=='#')
                q.push({i,j,0});
        }
    }
    bfs();
    cout<<dmax;
    return 0;
}