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
#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
#include<assert.h>
#include<numeric>
using namespace std;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
using ll = long long;
const int inf=1e9+7;
const ll longinf=1LL<<60 ;
const ll mod=1e9+7 ;

using ld = double;
using Point =  complex<ld>;
const ld eps = 1e-7;
const ld pi = acos(-1.0);

bool eq(ld a, ld b) {
	return (abs(a - b) < eps);
}
//内積
ld dot(Point a, Point b) {
	return real(conj(a)*b);
}
//外積
ld cross(Point a, Point b) {
	return imag(conj(a)*b);
}
//線分
//直線にするなら十分通い２点を端点とすればよい
class Line {
public:
	Point a, b;
};
//円
class Circle {
public:
	Point p;
	ld r;
};
//3点の位置関係
int ccw(Point a, Point b, Point c) {
	b -= a; c -= a;
	if (cross(b, c) > eps)return 1;//a,b,cが反時計回り
	if (cross(b, c) < -eps)return -1;//a,b,cが時計回り
	if (dot(b, c) < 0)return 2;//c,a,bの順に一直線
	if (norm(b) < norm(c))return -2;//a,b,cの順に一直線
	return 0;//a,c,bの順に一直線
}
//2直線の交差判定
bool isis_ll(Line l, Line m) {
	return !eq(cross(l.b - l.a, m.b - m.a), 0);
}
//直線と線分の交差判定
bool isis_ls(Line l, Line s) {
	return (cross(l.b - l.a, s.a - l.a)*cross(l.b - l.a, s.b - l.a) < eps);
}
//点が直線上に存在するか
bool isis_lp(Line l, Point p) {
	return (abs(cross(l.b - p, l.a - p)) < eps);
}
//点が線分上に存在するか
bool isis_sp(Line s, Point p) {
	return (abs(s.a - p) + abs(s.b - p) - abs(s.b - s.a) < eps);
}
//線分と線分の交差判定
bool isis_ss(Line s, Line t) {
	if (isis_sp(s, t.a) || isis_sp(s, t.b) || isis_sp(t, s.a) || isis_sp(t, s.b))return true;
	return(cross(s.b - s.a, t.a - s.a)*cross(s.b - s.a, t.b - s.a) < -eps && cross(t.b - t.a, s.a - t.a)*cross(t.b - t.a, s.b - t.a) < -eps);
}
//点から直線への垂線の足
Point proj(Line l, Point p) {
	ld t = dot(p - l.a, l.a - l.b) / norm(l.a - l.b);
	return l.a + t * (l.a - l.b);
}
//直線と直線の交点
//平行な２直線に対しては使うな！！！！
Point is_ll(Line s, Line t) {
	Point sv = s.b - s.a; Point tv = t.b - t.a;
	return s.a + sv * cross(tv, t.a - s.a) / cross(tv, sv);
}
//直線と点の距離
ld dist_lp(Line l, Point p) {
	return abs(p - proj(l, p));
}
//直線と直線の距離
ld dist_ll(Line l, Line m) {
	return isis_ll(l, m) ? 0 : dist_lp(l, m.a);
}
//線分と直線の距離
ld dist_ls(Line l, Line s) {
	return isis_ls(l, s) ? 0 : min(dist_lp(l, s.a), dist_lp(l, s.b));
}
//線分と点の距離
ld dist_sp(Line s, Point p) {
	Point r = proj(s, p);
	return isis_sp(s, r) ? abs(p - r) : min(abs(p - s.a), abs(p - s.b));
}
//線分と線分の距離
ld dist_ss(Line s, Line t) {
	if (isis_ss(s, t))return 0;
	return min({ dist_sp(s,t.a),dist_sp(s,t.b),dist_sp(t,s.a),dist_sp(t,s.b) });
}
int main(){
    int n;
    cin>>n;
    Point a[n+1];
    rep(i,n){
        double x,y;
        cin>>x>>y;
        a[i]={x,y};
    }
    if(n%2){
        cout<<"NA"<<endl;
        return 0;
    }
    a[n]=a[0];
    rep(i,n/2){
        if(abs(cross((a[i+1]-a[i]),a[i+n/2]-a[i+1+n/2]))>eps){
            cout<<"NA"<<endl;
            return 0;
        }
    }
    Point ans=(a[0]+a[n/2])*.5;
    cout<<setprecision(10)<<ans.real()<<" "<<ans.imag()<<endl;

    return 0;
}
