Problem:
Score : 300 points

Problem Statement

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.

Sample Input 1

3 3
...
.#.
...

Sample Output 1

2

After one operation, all but the corners of the grid will be black. After one more operation, all the squares will be black.

Sample Input 2

6 6
..#..#
......
#..#..
......
.#....
....#.

Sample Output 2

3

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include "bits/stdc++.h"

using namespace std;
using ll = long long int;
using ullong = unsigned long long;
#define rep(i,n) for(int i = 0; i < n; i++)
#define FOR(i, a, b)  for(int i = (a); i < (b) ; i++)
#define pb push_back
#define SORT(v,n) sort(v, v+n)
#define ALL(x) (x).begin(),(x).end()
#define debug(x) cerr << #x << ": " << x << '\n'
#define int ll
const int INF = 1e9;
const int MOD = (int)1e9 + 7;
const double EPS = 1e-9;
typedef pair<int, int> Vec2;
const int dx[8] = { 1, 0, -1, 0, 1, -1, -1, 1 };
const int dy[8] = { 0, 1, 0, -1, 1, 1, -1, -1 };


int h, w;
int d[1000][1000];
vector<string> grid(1000);

void printd() {
	cout << endl;
	rep(i, h) {
		rep(j, w) {
			cout << d[j][i];
		}
		cout << endl;
	}
	cout << endl;
}

void bfs(int x, int y) {
	queue<Vec2> Q;
	Q.push({ x,y });
	d[x][y] = 0;
	while (!Q.empty()) {
		int x = Q.front().first;
		int y = Q.front().second;
		Q.pop();
		rep(i, 4) {
			int nx = x + dx[i];
			int ny = y + dy[i];
			if (nx >= 0 and nx < w and ny >= 0 and ny < h) {
				if(d[x][y] + 1 < d[nx][ny]) {
				//if(d[x][y] == INF){
					Q.push({ nx,ny });
					d[nx][ny] = d[x][y] + 1;
				}
			}
		}
	}
}

signed main() {
	ios::sync_with_stdio(false);
	cin >> h >> w;
	vector<Vec2> vec;
	rep(i, h) {
		cin >> grid[i];
	}
	rep(i, 1000) {
		rep(j, 1000)
			d[i][j] = INF;
	}

	rep(i, h) {
		rep(j, w) {
			if (grid[i][j] == '#') {
				bfs(j, i);
				//printd();
			}
		}
	}

	int ans = 0;
	rep(i, h) {
		rep(j, w) {
			ans = max(ans, d[j][i]);
		}
	}
	cout << ans << endl;
	return 0;
}