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 <bits/stdc++.h>
using namespace std;

int vis[1005][1005];

int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  int n, m;
  cin >> n >> m;
  queue<array<int, 3>> q;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      char c;
      cin >> c;
      if (c == '#') {
        q.push({i, j, 0});
      }
    }
  }
  int sol = 0;
  while (q.size()) {
    int x = q.front()[0];
    int y = q.front()[1];
    int z = q.front()[2];
    q.pop();
    if (!x || !y || x > n || y > m) continue;
    if (vis[x][y]) continue;
    q.push({x + 1, y, z + 1});
    q.push({x - 1, y, z + 1});
    q.push({x, y + 1, z + 1});
    q.push({x, y - 1, z + 1});
    vis[x][y] = 1;
    sol = z;
  }
  cout << sol << '\n';
}