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, m, n) for (int i = m; i < n; ++i)

int dy[4] = {0, 1, 0, -1};
int dx[4] = {1, 0, -1, 0};

int main() {
  int H, W;
  cin >> H >> W;
  vector<string> A(H);
  rep(i, 0, H) cin >> A[i];
  using tp = tuple<int, int, int>;
  queue<tp> Q;
  rep(i, 0, H) rep(j, 0, W) {
    if (A[i][j] == '#') Q.push(tp(i, j, 0));
  }
  int ans = 0;
  while (!Q.empty()) {
    tp cur = Q.front();
    Q.pop();
    int d = get<2>(cur);
    rep(i, 0, 4) {
      int ny = get<0>(cur) + dy[i];
      int nx = get<1>(cur) + dx[i];
      if (ny < 0 || ny >= H || nx < 0 || nx >= W)
        continue;
      if (A[ny][nx] != '#') {
        A[ny][nx] = '#';
        Q.push(tp(ny, nx, d + 1));
        ans = max(ans, d + 1);
      }
    }
  }
  cout << ans << endl;
  return 0;
}
