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<bits/stdc++.h>
#define range(i,a,b) for(int i = (a); i < (b); i++)
#define rep(i,b) for(int i = 0; i < (b); i++)
#define all(a) (a).begin(), (a).end()
#define show(x)  cerr << #x << " = " << (x) << endl;
//const int INF = 1e8;
using namespace std;

int n;
char c[200][200];
const int dy[16] = { 0,-1, 0, 1, 1,-1, 1,-1, 0,-2, 0, 2};
const int dx[16] = { 1, 0,-1, 0, 1, 1,-1,-1, 2, 0,-2, 0};

void dfs(int x, int y){
	if(c[y][x] == 'o'){
		c[y][x] = 'x';
		rep(i,4){
			int nx = x + dx[i];
			int ny = y + dy[i];
			if(ny < 0 || ny >= n || nx < 0 || nx >= n) continue;
			dfs(nx, ny);
		}
	}
}

int main(){
	cin >> n;

	rep(i,n){
		rep(j,n){
			cin >> c[i][j];
		}
	}

	int cnt = 0;
	rep(i,n){
		rep(j,n){
			if(c[i][j] == 'o'){
				cnt++;
				dfs(j, i);
			}
		}
	}
	//show(cnt)
	cout << cnt / 3 << endl;
}