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>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using  namespace std;
typedef pair<int,int> P;

int dis[1005][1005];
int h, w;
int mx[] = {-1, 0, 1, 0};
int my[] = {0, -1, 0, 1};
queue<P> que;

void bfs() {
  while (!que.empty()) {
    P tmp = que.front(); que.pop();
    int x, y;
    x = tmp.first; y = tmp.second;
    rep(i, 4) {
      int tx, ty;
      tx = x + mx[i]; ty = y + my[i];
      if (dis[tx][ty] < 0) {
        dis[tx][ty] = dis[x][y] + 1;
        que.push(P (tx,ty));
      }
    }
  }
}

int main() {
  cin >> h >> w;
  rep(i, h) {
    rep(j, w) {
      char c;
      cin >> c;
      if (c == '#') que.push(P (i+1, j+1));
      else dis[i+1][j+1] = -1;
    }
  }
  bfs();
  int ans = 0;
  rep(i, h)rep(j, w) ans = max(ans, dis[i+1][j+1]);
  cout << ans << endl;
  return 0;
}