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>
using namespace std;
const int dx[]={1,-1,0,0};
const int dy[]={0,0,-1,1};
int n,m;
char a[1010][1010];
int ans[1010][1010];
vector<pair<int,int> > v;
int main(){
	memset(ans,0,sizeof(ans));
	cin>>n>>m;
	for(int i=0;i<n;i++){
		for(int j=0;j<m;j++){
			cin>>a[i][j];
			if(a[i][j]=='#')v.push_back(make_pair(i,j));
		}
	}
	int front=0;
	int rear=v.size();
	int cnt=0;
	while(front<rear){
		int x=v[front].first;
		int y=v[front].second;
		cnt=ans[x][y];
		for(int i=0;i<4;i++){
			int ax=x+dx[i];
			int ay=y+dy[i];
			if(ax>=0&&ax<n&&ay>=0&&ay<m&&a[ax][ay]=='.'){
				ans[ax][ay]=ans[x][y]+1;
				a[ax][ay]='#';
				v.push_back(make_pair(ax,ay));
				rear++;
			}
		}
		front++;
	}
	cout<<cnt<<endl;
	return 0;
}