Problem: A convex polygon consisting of N vertices is given. The coordinates of each vertex are represented counterclockwise by (X1, Y1), (X2, Y2), ……, (XN, YN). No matter what straight line passes through the point P, find the coordinates of the point P so that the areas of the two convex polygons obtained after cutting are equal.

Constraints

* All inputs are integers

* 3 ≤ N ≤ 50

* 0 ≤ | Xi |, | Yi | ≤ 1000000

* The input polygon is a simple convex polygon.

* The output must satisfy max (| X-cX |, | Y-cY |) ≤ 0.0001 when the output coordinates are (X, Y) and the exact solution is (cX, cY).

Input

The input is given in the following format.

> N
> X1 Y1
> X2 Y2
> ……
> XN YN
>

Output

If there is a point that satisfies the condition of the problem statement, the coordinates of that point

> X Y
>

Output in the format of. If the point does not exist, output "NA" on one line.

Examples

Input

4
100 100
0 100
0 0
100 0


Output

50.00000 50.00000


Input

3
100 100
0 100
0 0


Output

NA

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
#define EPS 1e-10
#define equals(a,b) (fabs( (a) - (b) )< EPS )
// c++ 11,14
#define COUNTER_CLOCKWISE 1
#define CLOCKWISE -1
#define ONLINE_BACK 2
#define ONLINE_FRONT -2
#define ON_SEGMENT 0

typedef struct point{
	double x,y;
	point(){};
	point(double x ,double y):x(x),y(y){};
	point operator + (point &p){ return point(x+p.x,y+p.y);	}
	point operator - (point &p){ return point(x-p.x,y-p.y);	}
	point operator * (point &p){ return point(x*p.x-y*p.y,x*p.y+y*p.x) ;}
	point operator * (double a){ return point(x*a,y*a);	}
	point operator / (double a){ return point(x/a,y/a);	}
	
	double abs() { return sqrt(norm()); }
	double norm() { return x*x+y*y; }
	
	bool operator < (const point &p) const { return x!=p.x ? x<p.x : y<p.y; }
	bool operator == (const point &p) const { return fabs(x-p.x)<EPS && fabs(y-p.y)<EPS; }
}point;
double abs(point a){return a.abs();}
double norm(point a){return a.norm();}

typedef complex<double> C;
typedef struct { 
	point s,e;
} line;

C convert(point a){ return C(a.x,a.y); }
point convert( C a){ return point(a.real(),a.imag() );}
double dot(point a,point b){ return a.x*b.x+a.y*b.y ; } //内積　a・b
double cross(point a,point b){ return a.x*b.y - a.y*b.x ; }//外積(z成分)　a×b
point vec(line l){return l.e-l.s;}
line make(point s,point e){
	line res; res.s=s; res.e=e;
	return res;
}
point make(){ 
	double x,y; cin>>x>>y;
       	return point(x,y);
}
line lmake(){
	point p0=make();
	point p1=make();
	return make(p0,p1);
}
//直交
bool isorthogonal(point a,point b){ return equals(dot(a,b), 0.0); }
bool isorthogonal(line l1,line l2){ return isorthogonal(vec(l1),vec(l2)); }
//平行
bool isparallel(point a,point b){ return equals(cross(a,b),0.0); }
bool isparallel(line l1,line l2){ return isparallel(vec(l1),vec(l2)); }
//射影
point project(line s,point p){
	point base = vec(s);
	double r=dot(p-s.s,base)/base.norm();
	base = base*r;
	return s.s+base;
}
//反射
point reflect(line l,point p){
	point tmp=project(l,p)-p;
	tmp= tmp*2.0;
	return p+tmp;

}
//交差判定

int ccw(point p0,point p1,point p2){
	point a = p1-p0;
	point b = p2-p0;
if(cross(a,b)>EPS) return 1;//counter_clockwise
if(cross(a,b)<-EPS) return -1;//clockwise
if(dot(a,b)<-EPS)return 2;//online_back
if(a.norm()<b.norm() ) return -2;//online_front
return 0;//on_segment
}
// line p1-p2 line p3-p4
bool intersect(point p1,point p2,point p3,point p4){
	return (ccw(p1,p2,p3)*ccw(p1,p2,p4)<=0 && ccw(p3,p4,p1)*ccw(p3,p4,p2)<=0 ); 
}
// line l1,l2
bool intersect(line l1,line l2){ return intersect(l1.s,l1.e,l2.s,l2.e); }


//距離
//point-point
double distance(point a,point b){ return abs(a-b); }
//point-line(直線）
double distance2(line l,point p){
	return abs(cross(vec(l),p-l.s)/abs(l.e-l.s));
}
//point-line(線分)
double distance(line l,point p){
	if( dot(vec(l),p-l.s) <0.0 ) return abs(p-l.s);
	if( dot(l.s-l.e,p-l.e) <0.0 ) return abs(p-l.e);
	return distance2(l,p);
}
//line-line
double distance(line l1,line l2){
	if(intersect(l1,l2))	return 0.0;
	return min(min(distance(l1,l2.s),distance(l1,l2.e) ), min(distance(l2,l1.s),distance(l2,l1.e) ) );
}

//交点
point crosspoint(line l1,line l2){
	point base = vec(l2);
	double d1 = abs(cross(base,l1.s-l2.s) );
	double d2 = abs(cross(base,l1.e-l2.s) );
	double t = d1/(d1+d2);
	point tmp = vec(l1)*t;
	return l1.s+tmp;
}
//面積
double area(vector<point> &p,int n){
	double ans=0.0;
	for(int i=0;i<n-2;i++){
	ans+=cross(p[i+2]-p[0],p[i+1]-p[0]);
	}
	return abs(ans)/2;
}
double area(vector<point> p){
	return area(p,p.size());
}
//内包
int contains(vector<point> &g,point p){
int n=g.size();
bool flag=false;
	for(int i=0;i<n;i++){
		point a=g[i]-p, b=g[(i+1)%n]-p;
		if(abs(cross(a,b) )<EPS && dot(a,b) <EPS)return 1;
		if(a.y>b.y) swap(a,b);
		if(a.y<EPS&&EPS<b.y&&cross(a,b)>EPS) flag=!flag;
	}
	return (flag? 2:0);
}
//凸包
vector<point> andrewscan(vector<point> &s){
	vector<point> u,l;
	if(s.size() <3) return s;
	sort(s.begin(),s.end());
	u.push_back(s[0]);
	u.push_back(s[1]);
	l.push_back(s[s.size()-1]);
	l.push_back(s[s.size()-2]);

	for(int i=2;i<s.size();i++){
			//凸包の辺上を含めない場合
		for(int n=u.size();n>=2&&ccw(u[n-2],u[n-1],s[i])!=CLOCKWISE;n--)
			//凸包の辺上を含める場合
		//for(int n=u.size();n>=2&&ccw(u[n-2],u[n-1],s[i])==COUNTER_CLOCKWISE;n--)
		{
			u.pop_back();
		}
		u.push_back(s[i]);
	}

	for(int i=s.size()-3;i>=0;i--){
			//凸包の辺上を含めない場合
		for(int n=l.size();n>=2&&ccw(l[n-2],l[n-1],s[i])!=CLOCKWISE;n--)
			//凸包の辺上を含める場合
		//for(int n=l.size();n>=2&&ccw(l[n-2],l[n-1],s[i])==COUNTER_CLOCKWISE;n--)
		{
			l.pop_back();
		}
		l.push_back(s[i]);
	}

	reverse(l.begin(),l.end());
	for(int i=u.size()-2;i>=1;i--)l.push_back(u[i]);
	return l;
}
//直径
double convex_diameter(vector<point> &g){
	int n=g.size();
	int is=0,js=0;
	for(int i=1;i<n;i++){
		if(g[is].y < g[i].y)is =i;
		if(g[i].y < g[js].y)js=i;
	}
	double maxd = abs((g[is]-g[js]));
	int i,mi,j,mj;
	i=mi=is;
	j=mj=js;
	do{
		if(cross(g[(i+1)%n]-g[i],g[(j+1)%n]-g[j])>=0)j=(j+1)%n;
		else i=(i+1)%n;
		if(abs(g[i]-g[j])>maxd){
			//cerr<<i<<' '<<j<<' '<<norm(g[i]-g[j])<<endl;
			maxd =abs(g[i]-g[j]);
			mi=i;mj=j;
		}
	}while(i!=is||j!=js);
	return maxd;
}
line half_line(vector<point> &g, int t){
	point p = g[t];
	double half = area(g)/2.0;
	double sum = 0.0;
	int i;
	double s;
	int n = g.size();
	for(i=0;i<g.size();i++){
		 s = cross(g[(i+t+1)%n]-p,g[(i+t+2)%n]-p)/2.0;
		if(s+sum+EPS>half)break;
		else sum+= s;
	}
	//cout<<t<<": "<<i+t+1<<' '<<half-sum<<' '<<s+sum-half<<endl;
	point a= g[(i+t+1)%n]*(s-half+sum) ,b= g[(i+t+2)%n]*(half-sum);
	point p2 = a+b;
	p2 = p2/s;
	return  make(p,p2);
}
point err = point(1e9,1e9);
point solve(vector<point> &g){
	double S =  area(g);
	double half = S/2.0;
	int n = g.size();
	vector<line>  l(n);
	for(int i= 0;i<n;i++) l[i] = half_line(g,i);
	vector<point> ps(n);
	bool f = false;
	for(int i=0;i<n;i++) ps[i] =  crosspoint(l[i],l[(i+1)%n] );
	//for(int i=0;i<n;i++) cout<<i<<":  "<<l[i].e.x<<' '<<l[i].e.y<<endl;
	//for(int i=0;i<n;i++) printf("%d: %0.6lf %0.6lf\n",i,ps[i].x,ps[i].y);
	for(int i=0;i<n;i++)if(abs(ps[i]-ps[(i+1)%n] ) <1e-6 ){

	}else{
		return err;
	}

	return ps[0];
}

int main(){
	int n;
	cin>>n;
	vector<point> g(n);
	for(int i=0;i<n;i++) g[i] = make();
	point ans = solve(g);
	if(ans == err||n==3){
		cout<<"NA"<<endl;
	}else{
		//cout<<ans.x<<' '<<ans.y<<endl;
		printf("%0.6lf %0.6lf\n",ans.x,ans.y);
	}
	return 0;
}