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
#include <iostream>
#include <vector>
#include <deque>
using namespace std;

struct pt{
	int x,y,k;
};

vector<vector<bool>> v(1001,vector<bool> (1001));
int h,w,ret=0;
deque<pt> q;
vector<pair<int,int>> dir={{0,1},{1,0},{-1,0},{0,-1}};

int bfs(){
	while (!q.empty()){
		auto f=q.front();
		ret=max(ret,f.k);
		q.pop_front();
		for (auto& i:dir){
			int nx=i.first+f.x,ny=f.y+i.second;
			if (nx < h and ny < w and nx >= 0 and ny >= 0 and !v[nx][ny]){
				v[nx][ny]=true;
				q.emplace_back(pt{nx,ny,f.k+1});
			}
		}
	}
	return ret;
}

int main(){
	char c;
	cin >> h >> w;
	for (int i=0; i<h; ++i){
		for (int j=0; j<w; ++j){
			cin >> c;
			if (c == '#'){
				q.emplace_back(pt{i,j,0});
				v[i][j]=true;
			}
		}
	}
	cout << bfs();
	return 0;
}