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 < (int)(n); i++)

int main() {
  using pii = pair<int,int>;
  int H, W;
  cin >> H >> W;
  char A[H][W];
  queue<pii> que;
  int d[H][W];
  rep(i, H) {
    rep(j, W) {
      cin >> A[i][j];
      if(A[i][j]=='#') que.push(pii(i,j));
      d[i][j]=0;
    }
  }
  int dy[4]={-1,0,1,0},dx[4]={0,1,0,-1};
  int ans=0;
  while(que.size()) {
    pii p=que.front();
    que.pop();
    rep(i,4) {
      int ny=p.first+dy[i], nx=p.second+dx[i];
      if(0<=ny&&ny<H&&0<=nx&&nx<W&&A[ny][nx]=='.'&&d[ny][nx]==0) {
        que.push(pii(ny,nx));
        d[ny][nx]=d[p.first][p.second]+1;
        ans=max(ans, d[ny][nx]);
      }
    }
  }
  cout << ans << endl;
}