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);

        // height
        int h = sc.nextInt();

        // width
        int w = sc.nextInt();

        // マス目の状態
        char[][] masu = new char[h][w];

        // BFSのQueueを準備
        Deque<XY> q = new ArrayDeque<>();

        // 入力を受け取る & '#'のマスをEnQueue
        for (int i = 0; i < h; i++) {
            masu[i] = sc.next().toCharArray();
            for (int j = 0; j < w; j++) {
                if ('#' == masu[i][j]) {
                    q.add(new XY(i, j, 0));
                }
            }
        }

        // ans
        int max_depth = 0;

        // BFS主処理
        for (;;) {
            // 無限ループ開始

            // 終了条件
            if (q.size() < 1) {
                break;
            }

            // queueからdequeue
            // pollメソッドはqueueから取り出し & queueから削除
            XY xy = q.poll();
            int x = xy.x;
            int y = xy.y;
            int depth = xy.depth;

            // 操作の回数を記録
            if (depth >= max_depth) {
                max_depth = depth;
            }

            // 現在の位置の周囲を黒にする
            if (x + 1 < w && masu[y][x + 1] == '.') {
                // 右マス
                masu[y][x + 1] = '#';
                q.add(new XY(y, x + 1, depth + 1));
            }
            if (x - 1 >= 0 && masu[y][x - 1] == '.') {
                // 左マス
                masu[y][x - 1] = '#';
                q.add(new XY(y, x - 1, depth + 1));
            }
            if (y + 1 < h && masu[y + 1][x] == '.') {
                // 下マス
                masu[y + 1][x] = '#';
                q.add(new XY(y + 1, x, depth + 1));
            }
            if (y - 1 >= 0 && masu[y - 1][x] == '.') {
                // 上マス
                masu[y - 1][x] = '#';
                q.add(new XY(y - 1, x, depth + 1));
            }

        }

        System.out.println(max_depth);

    }
}

class XY {
    int y;
    int x;

    int depth;

    XY(int y, int x, int d) {
        this.x = x;
        this.y = y;
        depth = d;
    }
}