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;
#define rep(i,n) for(int i=0;i<(n);i++)
typedef pair<int,int> P;
const int INF=1e9;
int main(){
    int h,w,ans=0;
    cin>>h>>w;
    vector<vector<char>> a(h,vector<char>(w));
    vector<vector<int>> dist(h,vector<int>(w,INF));
    queue<P> que;
    rep(i,h) rep(j,w){
        cin>>a[i][j];
        if(a[i][j]=='#'){
            que.push(P(i,j));
            dist[i][j]=0;
        }
    }
    const int dx[]={1,0,-1,0},dy[]={0,1,0,-1};
    while(que.size()){
        P p=que.front();que.pop();
        rep(i,4){
            int nx=p.first+dx[i],ny=p.second+dy[i];
            if(0<=nx&&nx<h&&0<=ny&&ny<w&&dist[nx][ny]==INF){
                que.push(P(nx,ny));
                dist[nx][ny]=dist[p.first][p.second]+1;
                ans=dist[nx][ny];
            }
        }
    }
    cout<<ans;
}