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.ArrayDeque;
import java.util.Arrays;
import java.util.Queue;
import java.util.Scanner;

public class Main {

    char[][] field;
    boolean[][] visited;
    int[] dx = {-1, 0, 1, 0};
    int[] dy = {0, 1, 0, -1};
    Queue<Integer[]> q = new ArrayDeque<>();

    void run() {
        Scanner sc = new Scanner(System.in);
        int h = sc.nextInt();
        int w = sc.nextInt();
        field = new char[h][w];
        visited = new boolean[h][w];
        int bCount = 0;

        for (int i = 0; i < h; i++) {
            char[] chars = sc.next().toCharArray();
            for (int j = 0; j < w; j++) {
                field[i][j] = chars[j];
                if (chars[j] == '#') {
                    q.add(new Integer[]{i, j, 0});
                    visited[i][j] = true;
                }
            }
        }

        int count = -1;
        while (!q.isEmpty()) {
            Integer[] next = q.remove();
            int x = next[0];
            int y = next[1];
            int depth = next[2];
            count = Math.max(count, depth);
            for (int i = 0; i < 4; i++) {
                int nx = x + dx[i];
                int ny = y + dy[i];
                if (0 <= nx && nx < h && 0 <= ny && ny < w && !visited[nx][ny]) {
                    visited[nx][ny] = true;
                    q.add(new Integer[]{nx, ny, depth + 1});
                }
            }
        }

        System.out.println(count);
    }

    void debug(Object... os) {
        System.err.println(Arrays.deepToString(os));
    }

    public static void main(String[] args) {
        new Main().run();
    }

}