Problem: There is a square-shaped grid with N vertical rows and N horizontal columns. We will denote the square at the i-th row from the top and the j-th column from the left as (i,\ j).

Initially, each square is either white or black. The initial color of the grid is given to you as characters a_{ij}, arranged in a square shape. If the square (i,\ j) is white, a_{ij} is `.`. If it is black, a_{ij} is `#`.

You are developing a robot that repaints the grid. It can repeatedly perform the following operation:

* Select two integers i, j (1 ≤ i,\ j ≤ N). Memorize the colors of the squares (i,\ 1), (i,\ 2), ..., (i,\ N) as c_1, c_2, ..., c_N, respectively. Then, repaint the squares (1,\ j), (2,\ j), ..., (N,\ j) with the colors c_1, c_2, ..., c_N, respectively.



Your objective is to turn all the squares black. Determine whether it is possible, and find the minimum necessary number of operations to achieve it if the answer is positive.

Constraints

* 2 ≤ N ≤ 500
* a_{ij} is either `.` or `#`.

Input

The input is given from Standard Input in the following format:


N
a_{11}...a_{1N}
:
a_{N1}...a_{NN}


Output

If it is possible to turn all the squares black, print the minimum necessary number of operations to achieve the objective. If it is impossible, print `-1` instead.

Examples

Input

2
#.
.#


Output

3


Input

2
.
.#


Output

3


Input

2
..
..


Output

-1


Input

2


Output

0


Input

3
.#.

.#.


Output

2


Input

3
...
.#.
...


Output

5

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
char stage[501][501];
int n;
int countblackcol(){
	int res=0;
	for (int x = 0; x < n; ++x)
	{
		int y;
		for (y = 0; y < n; ++y)
		{
			if(stage[y][x]=='.') break;
		}
		if(y==n) res++;
	}
	return res;
}
int searchrow(){
	//黒一行を作るまでの回数の最小を調べる
	int minv = n+1;
	for (int y = 0; y < n; ++y)
	{
		int v = 0;
		int ty;
		//塗りつぶし開始までの準備
		for (ty = 0; ty < n; ++ty)
		{
			if(stage[ty][y]=='#') break;
		}
		if(ty == n) v=1;
		//塗りつぶす個数
		for (int x = 0; x < n; ++x)
		{
			if(stage[y][x]=='.') v++;
		}
		//行動回数が最小のものを残す
		if(minv > v){
			minv = v;
		}
		
	}
	return minv;
}
int main(){
	// freopen("in.txt","r",stdin);
	cin >> n;
	for (int i = 0; i < n; ++i)
	{
		scanf("%s",stage[i]);
	}
	//どこに張り付けたとしても縦一列にはならない
	//つまり初めに縦一列黒なものから増えない
	int ans = searchrow();
	if(ans==n+1){
		cout << "-1" << endl;
		return 0;
	}
	//縦一列黒ではない個数を調べる
	ans += n - countblackcol();
	cout << ans << endl;
	//横一列すべて黒にするまでにかかる最小試行数
	

	return 0;
}