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.Deque;
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();
        char[][] a = new char[h][w];
        Deque<int[]> queue = new ArrayDeque<>();
        boolean[][] visited = new boolean[h][w];
        for (int i = 0; i < h; i++) {
            a[i] = sc.next().toCharArray();
            for (int j = 0; j < w; j++) {
                if (a[i][j] == '#') {
                    visited[i][j] = true;
                    queue.add(new int[] { i, j });
                }
            }
        }
        int cnt = -1;
        final int[][] DXDY = { { 1, 0 }, { 0, 1 }, { -1, 0 }, { 0, -1 } };
        while (!queue.isEmpty()) {
            int size = queue.size();
            cnt++;
            for (int i = 0; i < size; i++) {
                int[] p = queue.poll();
                for (int j = 0; j < DXDY.length; j++) {
                    int x = p[0] + DXDY[j][0];
                    int y = p[1] + DXDY[j][1];
                    if (x < 0 || x >= h || y < 0 || y >= w || visited[x][y]) {
                        continue;
                    }
                    visited[x][y] = true;
                    queue.add(new int[] { x, y });
                }
            }
        }
        System.out.println(cnt);
    }
}