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.awt.*;
import java.util.ArrayDeque;
import java.util.Queue;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int h = sc.nextInt();
        int w = sc.nextInt();
        int[][] arr = new int[h][w];
        Queue<Point> que = new ArrayDeque<>();

        for (int i = 0; i < h; i++) {
            String line = sc.next();
            for (int s = 0; s < w; s++) {
                if (line.charAt(s) == '#') {
                    que.add(new Point(s, i));
                    arr[i][s] = 1;
                }
            }
        }

        while (!que.isEmpty()) {
            Point p = que.poll();
            if (0 <= p.x - 1 && arr[p.y][p.x - 1] == 0) {
                arr[p.y][p.x - 1] = arr[p.y][p.x] + 1;
                que.add(new Point(p.x - 1, p.y));
            }
            if (p.x + 1 < w && arr[p.y][p.x + 1] == 0) {
                arr[p.y][p.x + 1] = arr[p.y][p.x] + 1;
                que.add(new Point(p.x + 1, p.y));
            }

            if (0 <= p.y - 1 && arr[p.y - 1][p.x] == 0) {
                arr[p.y - 1][p.x] = arr[p.y][p.x] + 1;
                que.add(new Point(p.x, p.y - 1));
            }
            if (p.y + 1 < h && arr[p.y + 1][p.x] == 0) {
                arr[p.y + 1][p.x] = arr[p.y][p.x] + 1;
                que.add(new Point(p.x, p.y + 1));
            }
        }

        int ans = 0;
        for (int[] a : arr) {
            for (int b : a) {
                ans = Math.max(ans, b);
            }
        }
        System.out.println(ans-1);
    }
}