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 main () {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int h, w;
  int dx[] = {1, 0, -1, 0};
  int dy[] = {0, 1, 0, -1};
  char c;
  queue<pair<int, int>> q;
  cin >> h >> w;
  vector<vector<int>> v(h+2, vector<int>(w+2, 0));
  for (int i = 1; i <= h; ++i) {
    for (int j = 1; j <=w; ++j) {
      cin >> c;
      if (c == '.') v[i][j] = -1;
      else q.emplace(i, j);
    }
  }
  while (!q.empty()) {
    int y = q.front().first;
    int x = q.front().second;
    q.pop();
    for (int i = 0; i < 4; ++i) {
      if (v[y + dy[i]][x + dx[i]] == -1) {
        v[y + dy[i]][x + dx[i]] = v[y][x] + 1;
        q.emplace(y + dy[i], x + dx[i]);
      }
    }
  }
  int mx = 0;
  for (int i = 1; i <= h; ++i) {
    for (int j = 1; j <= w; ++j) {
      mx = max(mx, v[i][j]);
    }
  }
  cout << mx << endl;
}