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;

const int dx[]={1,-1,0,0};
const int dy[]={0,0,1,-1};


char s[1015][1015];
struct node{ int x,y,step; };
queue<node> q;
int ans;

void bfs() {
    while (!q.empty()) {
        node now = q.front(); q.pop();
        ans=max(ans,now.step);
        for (int i=0;i<4;i++) {
            node nex = node{now.x+dx[i],now.y+dy[i],now.step+1};
            if (s[ nex.x ][ nex.y ]=='.') {
                s[ nex.x ][ nex.y ]='#';
                q.push(nex);
            }
        }
    }
}

int main () {
    int n,m;
    cin >> n >> m;
    
    for (int i=1;i<=n;i++)
        for (int j=1;j<=m;j++) {
            cin >> s[i][j];
            if (s[i][j]=='#')
                q.push(node{i,j,0});
        }
    
    bfs();
    cout << ans << endl;
    return 0;
}