Problem

There is a grid of $ N \ times N $ cells. Initially all cells are white. Follow the steps below to increase the number of black squares.

Select one white cell from the cells that are even-numbered from the top and even-numbered from the left. The selected square turns black. Further, the adjacent white squares also change to black in a chain reaction in each of the vertical and horizontal directions of the squares. This chain continues until there are no white cells in that direction. (An example is shown below)

□□□□□□□□□
□□□□□□□□□
□□□□□□□□□
□□□□□□□□□
□□□□□□□□□
□□□□□□□□□
□□□□□□□□□
□□□□□□□□□
□□□□□□□□□
↓ Select the 4th from the top and the 6th from the left
□□□□□ ■ □□□
□□□□□ ■ □□□
□□□□□ ■ □□□
■■■■■■■■■
□□□□□ ■ □□□
□□□□□ ■ □□□
□□□□□ ■ □□□
□□□□□ ■ □□□
□□□□□ ■ □□□
↓ Select the second from the top and the second from the left
□ ■ □□□ ■ □□□
■■■■■■ □□□
□ ■ □□□ ■ □□□
■■■■■■■■■
□□□□□ ■ □□□
□□□□□ ■ □□□
□□□□□ ■ □□□
□□□□□ ■ □□□
□□□□□ ■ □□□
↓ Select 6th from the top and 8th from the left
□ ■ □□□ ■ □□□
■■■■■■ □□□
□ ■ □□□ ■ □□□
■■■■■■■■■
□□□□□ ■ □ ■ □
□□□□□ ■■■■
□□□□□ ■ □ ■ □
□□□□□ ■ □ ■ □
□□□□□ ■ □ ■ □


You want to create a grid where the color of the $ i $ th cell from the top and the $ j $ th cell from the left is $ A_ {i, j} $. Find the minimum number of times to select a square.

Constraints

The input satisfies the following conditions.

* $ 3 \ le N \ le 127 $
* $ N $ is odd
* $ A_ {i, j} $ is'o'or'x'
* Only input that can be made is given

Input

The input is given in the following format.

$ N $
$ A_ {1,1} $ $ A_ {1,2} $ ... $ A_ {1, N} $
$ A_ {2,1} $ $ A_ {2,2} $ ... $ A_ {2, N} $
...
$ A_ {N, 1} $ $ A_ {N, 2} $ ... $ A_ {N, N} $


When $ A_ {i, j} $ is'o', it means that the $ i $ th cell from the top and the $ j $ th cell from the left are white, and when it is'x', it is a black cell. Represents that.

Output

Outputs the minimum number of times to select a cell on the first line.

Examples

Input

5
oxooo
oxooo
oxooo
xxxxx
oxooo


Output

1


Input

9
oxoooooxo
oxxxxxxxx
oxoooooxo
xxxxxxxxx
oxoxooooo
oxxxxxxxx
oxoxoxooo
oxoxxxxxx
oxoxoxooo


Output

4
#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<complex>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
using namespace std;

typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
const ll mod = 1000000007/4;
typedef double ld;
typedef complex<ld> Point;
const ll INF = mod * mod;
typedef pair<int, int> P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
const ld eps = 1e-8;
const ld pi = acos(-1.0);
typedef pair<ld, ld> LDP;
typedef pair<ll, ll> LP;
typedef vector<int> vec;
typedef vector<string> svec;

int n;
char mp[127][127];

map<pair<P, P>, bool> used;
map<pair<P, P>, int> ans;
int dfs(int lx, int ly, int rx, int ry) {
	pair<P, P> p = { {lx,ly},{rx,ry} };
	if (used[p])return ans[p];
	used[p] = true;
	bool f = true;
	Rep1(i, lx, rx) {
		Rep1(j, ly, ry) {
			if (mp[i][j] == 'x')f = false;
		}
	}
	if (f)return ans[p] = 0;
	vector<bool> h(rx - lx + 1, false);
	vector<bool> w(ry - ly + 1, false);
	int ret = mod;
	Rep1(i, lx, rx) {
		h[i-lx] = true;
		Rep1(j, ly, ry) {
			h[i-lx] = h[i-lx]&(mp[i][j] == 'x');
		}
	}
	Rep1(j, ly, ry) {
		w[j - ly] = true;
		Rep1(i, lx, rx) {
			w[j - ly] = w[j - ly] & (mp[i][j] == 'x');
		}
	}
	rep(i, rx - lx + 1) {
		if (!h[i])continue;
		rep(j, ry - ly + 1) {
			if (!w[j])continue;
			int x = i + lx;
			int y = j + ly;
			int sum = 1 + dfs(lx, ly, x - 1, y - 1) + dfs(x + 1, ly, rx, y - 1) + dfs(lx, y + 1, x - 1, ry) + dfs(x + 1, y + 1, rx, ry);
			ret = min(ret, sum);
		}
	}
	return ans[p] = ret;
}
void solve() {
	cin >> n;
	rep(i, n) {
		rep(j, n) {
			cin >> mp[i][j];
		}
	}
	cout << dfs(0, 0, n - 1, n - 1) << endl;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	//cout << fixed << setprecision(5);
	//while (cin >> n>>h>>m>>s,n)solve();
	solve();
	//stop
	return 0;
}

