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 CPP):

Solution:
#include <bits/stdc++.h>
#define rep(i, n) REP(i, 0, n)
#define REP(i, l, r) for (long long i = l; i < r; i++)
using namespace std;
int mx[4]={0,1,0,-1},my[4]={1,0,-1,0};
signed main() {
	int h,w,ans=0;
	char c[1000][1000];
	queue<pair<pair<int,int>,int>> que;
	cin>>h>>w;
	rep(i,h){
		rep(j,w){
			cin>>c[i][j];
			if(c[i][j]=='#')que.push({{i,j},0ll});
		}
	}
	while(!que.empty()){
		pair<pair<int,int>,int> tu=que.front();que.pop();
		rep(i,4){
			if(tu.first.first+mx[i]<0||tu.first.first+mx[i]>=h||tu.first.second+my[i]<0||tu.first.second+my[i]>=w)continue;
			if(c[tu.first.first+mx[i]][tu.first.second+my[i]]=='.'){
				que.push({{tu.first.first+mx[i],tu.first.second+my[i]},tu.second+1});
				ans=tu.second+1;
				c[tu.first.first+mx[i]][tu.first.second+my[i]]='#';
			}
		}
	}
	cout<<ans<<endl;
	return 0;
}