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 <iostream>
#include <string>
#include <queue>
#include <algorithm>
#include <utility>
using namespace std;

int a[1010][1010];
int dr[4]={-1, 0, 1, 0};
int dc[4]={0, -1, 0, 1};
int main()
{
    int n, m; cin>>n>>m;
    queue<pair<int, int>> q;
    for(int i=0; i<n; ++i) {
        string s; cin>>s;
        for(int j=0; j<m; ++j) if (s[j]=='#') {
            q.emplace(i, j);
            a[i][j]=1;
        }
    }

    int res=0;
    while (!q.empty()) {
        auto p=q.front(); q.pop();
        int r=p.first, c=p.second;
        res=max(res, a[r][c]-1);
        for(int i=0; i<4; ++i) {
            int nr=r+dr[i], nc=c+dc[i];
            if (nr<0 or nr>=n or nc<0 or nc>=m) continue;
            if (a[nr][nc]) continue;
            q.emplace(nr, nc);
            a[nr][nc]=a[r][c]+1;
        }
    }
    cout<<res<<endl;
}