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)
using ll = long long;
const ll MOD = 1000000007;//998244353;
const int INF = 10000;
int A[1000][1000];
int main() {
   ios::sync_with_stdio(false);
   cin.tie(0);
   int h, w; cin >> h >> w;
   rep(i,h){
      rep(j,w){
         char c; cin >> c;
         A[i][j] = c=='#' ? 0 : INF;
      }
   }
   rep(i,h){
      rep(j,w-1){
         A[i][j+1] = min(A[i][j]+1, A[i][j+1]);
         A[i][w-j-2] = min(A[i][w-j-1]+1, A[i][w-j-2]);
      }
   }
   rep(j,w){
      rep(i,h-1){
         A[i+1][j] = min(A[i][j]+1, A[i+1][j]);
         A[h-i-2][j] = min(A[h-i-1][j]+1, A[h-i-2][j]);
      }
   }
   int ans = 0;
   rep(i,h)rep(j,w)ans = max(ans, A[i][j]);
   cout << ans << endl;
}