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 static java.util.Comparator.*;

import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;
import java.util.TreeSet;

public class Main {
    public static void main(String[] args) {
        PrintWriter out = new PrintWriter(System.out);
        Solver solver = new Solver(System.in, out);
        solver.solve();
        out.close();
    }
}
class Solver {
	Scanner sc;
	PrintWriter out;
    public Solver(InputStream in, PrintWriter out) {
    	sc = new Scanner(in);
    	this.out = out;
    }
    // ==========================================================
    public void solve() {
    	int H = ni();
    	int W = ni();
    	char[][] map = new char[H][W];
    	Queue<PP> que1 = new ArrayDeque<>();
    	Queue<PP> que2 = new ArrayDeque<>();
    	for (int i = 0; i < H; i++) {
			map[i] = ns().toCharArray();
			for (int j = 0; j < W; j++) {
				if(map[i][j] == '#') {
					que1.add(new PP(i,j));
				}
			}
		}
    	int ans = 0, h, w;
    	PP p = null;
    	while(true) {
	    	que2.clear();
	    	while(!que1.isEmpty()) {
	    		p = que1.poll();
	    		h = p.key;
	    		w = p.val;
	    		if(h > 0 && map[h-1][w] == '.') {
	    			map[h-1][w] = '#';
	    			que2.add(new PP(h-1, w));
	    		}
	    		if(w > 0 && map[h][w-1] == '.') {
	    			map[h][w-1] = '#';
	    			que2.add(new PP(h, w-1));
	    		}
	    		if(w < W-1 && map[h][w+1] == '.') {
	    			map[h][w+1] = '#';
	    			que2.add(new PP(h, w+1));
	    		}
	    		if(h < H-1 && map[h+1][w] == '.') {
	    			map[h+1][w] = '#';
	    			que2.add(new PP(h+1, w));
	    		}
	    	}
	    	if(que2.isEmpty())	break;
	    	ans++;
	    	que1.addAll(que2);
    	}
    	out.println(ans);
    }
    // Set に入れるなら　PPKEY　を使う！
    static class PP{
    	public int key, val;
    	public PP(int key, int val) {
    		this.key = key;
    		this.val = val;
    	}
		public int getKey() {
			return key;
		}
		public void setKey(int key) {
			this.key = key;
		}
		public int getVal() {
			return val;
		}
		public void setVal(int val) {
			this.val = val;
		}
    }
    // ------------------------------------------
    // 入力
    // ------------------------------------------
    public int ni() {
    	return sc.nextInt();
    }
    public long nl() {
    	return sc.nextLong();
    }
    public String ns() {
    	return sc.next();
    }
    public char[] nc() {
    	return sc.next().toCharArray();
    }
    public int[] ndi(int N) {
    	int[] ans = new int[N];
    	for (int i = 0; i < N; i++) {
			ans[i] = ni();
		}
    	return ans;
    }
    public long[] ndl(int N) {
    	long[] ans = new long[N];
    	for (int i = 0; i < N; i++) {
			ans[i] = nl();
		}
    	return ans;
    }
    public String[] nds(int N) {
    	String[] ans = new String[N];
    	for (int i = 0; i < N; i++) {
			ans[i] = ns();
		}
    	return ans;
    }
    public char[][] ndc(int N) {
    	char[][] ans = new char[N][];
    	for (int i = 0; i < N; i++) {
			ans[i] = nc();
		}
    	return ans;
    }
    public int[][] nddi(int N, int S) {
    	int[][] ans = new int[N][S];
    	for (int i = 0; i < N; i++) {
			for (int j = 0; j < S; j++) {
				ans[i][j] = ni();
			}
		}
    	return ans;
    }
}