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 JAVA):

Solution:
import java.util.*;

class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int h = Integer.parseInt(sc.next());
        int w = Integer.parseInt(sc.next());
        int s[][] = new int[h + 2][w + 2];
        Queue<Pair> que = new ArrayDeque<>();
        for (int i = 1; i <= h; i++) {
            char c[] = sc.next().toCharArray();
            for (int j = 1; j <= w; j++) {
                if (c[j - 1] == '.') {
                    s[i][j] = -1;
                } else {
                    que.add(new Pair(i, j));
                }

            }
        }
        sc.close();

        Pair p;
        int ans = 0;
        while (!que.isEmpty()) {
            p = que.poll();
            ans = s[p.x][p.y];
            if (s[p.x + 1][p.y] == -1) {
                s[p.x + 1][p.y] = ans + 1;
                que.add(new Pair(p.x + 1, p.y));
            }
            if (s[p.x - 1][p.y] == -1) {
                s[p.x - 1][p.y] = ans + 1;
                que.add(new Pair(p.x - 1, p.y));
            }
            if (s[p.x][p.y + 1] == -1) {
                s[p.x][p.y + 1] = ans + 1;
                que.add(new Pair(p.x, p.y + 1));
            }
            if (s[p.x][p.y - 1] == -1) {
                s[p.x][p.y - 1] = ans + 1;
                que.add(new Pair(p.x, p.y - 1));
            }
        }
        System.out.println(ans);
    }
}

class Pair {
    int x;
    int y;

    public Pair(int x, int y) {
        this.x = x;
        this.y = y;
    }
}