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.*;

public class Main {
	static int[][] DIR = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
	public static final int Y = 1024;

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int H = sc.nextInt();
		int W = sc.nextInt();
		char[][] A = new char[H][];
		int num_w = 0;
		int num_b = 0;

		HashSet<Integer> black = new HashSet<>();

		for (int i = 0; i < H; i++) {
			String s = sc.next();
			char[] cs = s.toCharArray();
			A[i] = cs;
			for (int j = 0; j < cs.length; j++) {
				if (cs[j] == '.') {
					num_w ++;
				}
				else {
					black.add(i * Y + j);
					num_b++;
				}
			}
		}
		int turn = 0;
		while (num_w > 0) {
			turn++;

			HashSet<Integer> black2 = new HashSet<>();
			for (int pair : black) {
				int _y = pair / Y;
				int _x = pair % Y;
				for (int d = 0; d < DIR.length; d++) {
					int x = _x + DIR[d][0];
					int y = _y + DIR[d][1];

					if (0 <= y && y < A.length && 0 <= x && x < A[y].length && A[y][x] == '.') {
						A[y][x] = '#';
						black2.add(y * Y + x);
						num_w--;
					}
				}
			}
			black = black2;
		}
		System.out.println(turn);
	}
}