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;

signed main() {
  ios::sync_with_stdio(false);

  int N, M;
  cin >> N >> M;

  vector<string> G(N);
  for (int i = 0; i < N; ++i) cin >> G[i];

  queue<tuple<int, int, int>> que;
  for (int i = 0; i < N; ++i) {
    for (int j = 0; j < M; ++j) {
      if (G[i][j] == '#') que.emplace(0, i, j);
    }
  }

  int ans = 0;
  while (que.size()) {
    int t, x, y;
    tie(t, x, y) = que.front();
    que.pop();
    ans = max(ans, t);
    for (int di = 0; di < 4; ++di) {
      const int dx[] = {1, 0, -1, 0};
      const int dy[] = {0, 1, 0, -1};
      int nx = x + dx[di];
      int ny = y + dy[di];
      if (nx < 0 || nx == N || ny < 0 || ny == M) continue;
      if (G[nx][ny] == '#') continue;
      G[nx][ny] = '#';
      que.emplace(t + 1, nx, ny);
    }
  }

  cout << ans << endl;
  
  return 0;
}
