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>
#define ll long long 
#define P pair<int,int>
#define PP pair<P,int>
using namespace std;
int inf = 1ll<<30;
int dx[4] = {1,0,-1,0};
int dy[4] = {0,1,0,-1};

int main(){
    int h,w;
    cin >> h >> w;
    char g[1005][1005];
    queue<PP>que;
    for(int i=0;i<h;i++)for(int j=0;j<w;j++) {
        cin >> g[i][j];
        if(g[i][j] == '#') que.push(PP{P{i, j}, 0});
    }

    int ans;
    while(!que.empty()){
        PP p = que.front(); que.pop();
        ans = p.second;
        for(int i=0;i<4;i++){
            int nx = p.first.first + dx[i];
            int ny = p.first.second + dy[i];
            if(!(0 <= nx && 0 <= ny && h > nx && w > ny)) continue;
            if(g[nx][ny] == '#') continue;
            g[nx][ny] = '#';
            que.push(PP{P{nx, ny}, ans + 1});
        }
    }

    cout << ans << endl;

}