Problem:
Score : 1300 points

Problem Statement

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 #.

Partial Score

In a test set worth 300 points, N ≤ 3.

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.

Sample Input 1

2
#.
.#

Sample Output 1

3

For example, perform the operation as follows:

Select i = 1, j = 2.

Select i = 1, j = 1.

Select i = 1, j = 2.

The transition of the colors of the squares is shown in the figure below:

Sample Input 2

2
..
..

Sample Output 2

-1

Sample Input 3

2
##
##

Sample Output 3

0

Sample Input 4

3
.#.
###
.#.

Sample Output 4

2

Sample Input 5

3
...
.#.
...

Sample Output 5

5

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#pragma GCC optimize("Ofast", "unroll-loops")

#include <bits/stdc++.h>

using namespace std;

#define ll long long

const int inf = 1000000000;

int n;
vector<string> a;

void input() {
	cin >> n;
	a.resize(n);
	for (int i = 0; i < n; ++i)
		cin >> a[i];
}

// 盤面がすべて白かどうかを判定
bool all_white() {
	for (int i = 0; i < n; ++i)
		for (int j = 0; j < n; ++j)
			if (a[i][j] == '#')
				return false;
	return true;
}

// まず上からx行目をall blackにするとき、全体での回数
int solve(int x) {
	int res = 0;
	// 上からx行目をall blackにする
	bool b_exist = false;
	for (int j = 0; j < n; ++j) {
		if (a[j][x] == '#') {
			b_exist = true;
			break;
		}
	}
	for (int i = 0; i < n; ++i) {
		if (a[x][i] == '#') continue;
		++res;
	}
	if (res && (!b_exist)) ++res;
	// 左の列から順にall blackにする
	for (int j = 0; j < n; ++j) {
		bool all_black = true;
		for (int i = 0; i < n; ++i) {
			if (a[i][j] == '.') {
				all_black = false;
				break;
			}
		}
		if (!all_black) ++res;
	}
	return res;
}

int main() {
	input();
	if (all_white()) cout << -1 << endl;
	else {
		int ans = inf;
		for (int x = 0; x < n; ++x)
			ans = min(ans, solve(x));
		cout << ans << endl;
	}
	return 0;
}