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<vector>
#include<string>
#include<algorithm>	
#include<map>
#include<set>
#include<utility>
#include<cmath>
#include<cstring>
#include<queue>
#include<stack>
#include<cstdio>
#include<sstream>
#include<iomanip>
#include<assert.h>
#define loop(i,a,b) for(int i=a;i<b;i++) 
#define rep(i,a) loop(i,0,a)
#define pb push_back
#define all(in) in.begin(),in.end()
#define shosu(x) fixed<<setprecision(x)
using namespace std;
//kaewasuretyuui
typedef long long ll;
typedef int Def;
typedef pair<Def,Def> pii;
typedef vector<Def> vi;
typedef vector<vi> vvi;
typedef vector<pii> vp;
typedef vector<vp> vvp;
typedef vector<string> vs;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef pair<Def,pii> pip;
typedef vector<pip>vip;
//#define mt make_tuple
//typedef tuple<double,int,double> tp;
//typedef vector<tp> vt;
template<typename A,typename B>bool cmin(A &a,const B &b){return a>b?(a=b,true):false;}
template<typename A,typename B>bool cmax(A &a,const B &b){return a<b?(a=b,true):false;}
const double PI=acos(-1);
const double EPS=1e-7;
Def inf=sizeof(Def)==sizeof(long long)?2e18:1e9;
int dx[]={0,1,0,-1};
int dy[]={1,0,-1,0};
vs in;
int f(int a,int b,int c,int d){
//	cout<<a<<" "<<b<<" "<<c<<" "<<d<<endl;
	bool H=false;
	loop(i,a,c)loop(j,b,d)if(in[i][j]=='x')H=true;
	if(!H)return 0;
	vi ta,yo;
	loop(i,a,c){
		bool h=true;
		loop(j,b,d)if(in[i][j]=='o')h=false;
		if(h)yo.pb(i);
	}
	loop(i,b,d){
		bool h=true;
		loop(j,a,c)if(in[j][i]=='o')h=false;
		if(h)ta.pb(i);
	}
	int out=0;
//	cout<<a<<" "<<b<<" "<<c<<" "<<d<<endl;
//	cout<<yo.size()<<" "<<ta.size()<<endl;
	rep(i,yo.size())rep(j,ta.size())
		out=max(out,f(a,b,yo[i],ta[j])+f(a,ta[j]+1,yo[i],d)+f(yo[i]+1,b,c,ta[j])+f(yo[i]+1,ta[j]+1,c,d)+1);
	return out;
}
int main(){
	int n;
	cin>>n;
	in=vs(n);
	rep(i,n)cin>>in[i];
	cout<<f(0,0,n,n)<<endl;
}