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;

const double EPS = 1e-12;
const double PI = acos(-1.0);

inline double SQ(double a){
	return a * a;
}

inline bool EQ(double a, double b){
	return abs(a - b) < EPS;
}

inline bool LT(double a, double b){
	return a - b < -EPS;
}

inline bool LEQ(double a, double b){
	return a - b < EPS;
}

inline bool GT(double a, double b){
	return LT(b, a);
}

inline bool GEQ(double a, double b){
	return LEQ(b, a);
}

typedef complex<double> Point;

#define x() real()
#define y() imag()

namespace std {

bool operator < (const Point &a, const Point &b){
	return a.x() != b.x() ? a.x() < b.x() : a.y() < b.y();
}

inline istream& operator >> (istream &is, Point &p){
	double x, y;
	is >> x >> y;
	p = Point(x, y);
	return is;
}

inline ostream& operator << (ostream &os, const Point &p){
	return os << p.x() << " " << p.y();
}

}

struct Line {
	Point p1, p2;
};

typedef Line Segment;

struct Circle {
	Point p;
	double r;
};

typedef vector<Point> Polygon;

double norm(const Point &a){
	return SQ(a.x()) + SQ(a.y());
}

double dot(const Point &a, const Point &b){
	return (conj(a) * b).x();
}

double cross(const Point &a, const Point &b){
	return (conj(a) * b).y();
}

bool orthogonal(const Segment &s1, const Segment &s2){
	return EQ(dot(s1.p2 - s1.p1, s2.p2 - s2.p1), 0.0);
}

bool parallel(const Segment &s1, const Segment &s2){
	return EQ(cross(s1.p2 - s1.p1, s2.p2 - s2.p1), 0.0);
}

Point project(const Line &l, const Point &p){
	Point base = l.p2 - l.p1;
	double r = dot(p - l.p1, base) / norm(base);
	return l.p1 + base * r;
}

Point reflect(const Line &l, const Point &p){
	return p + (project(l, p) - p) * 2.0;
}

enum CCW {
	COUNTER_CLOCKWISE = 1,
	CLOCKWISE = -1,
	ONLINE_BACK = 2,
	ONLINE_FRONT = -2,
	ON_SEGMENT = 0
};

int ccw(const Point &p0, const Point &p1, const Point &p2){
	Point a = p1 - p0;
	Point b = p2 - p0;
	if (GT(cross(a, b), 0.0)) return COUNTER_CLOCKWISE;
	if (LT(cross(a, b), 0.0)) return CLOCKWISE;
	if (LT(dot(a, b), 0.0)) return ONLINE_BACK;
	if (LT(norm(a), norm(b))) return ONLINE_FRONT;
	return ON_SEGMENT;
}

// ????????§?????? or ????????´???
bool intersectLL(const Line &l, const Line &m){
	Point va = l.p2 - l.p1, vb = m.p2 - l.p1;
	return !parallel(l, m) || EQ(cross(va, vb), 0.0);
}

bool intersectLS(const Line &l, const Segment &s){
	Point va = l.p2 - l.p1, vb = s.p1 - l.p1, vc = s.p2 - l.p1;
	return LEQ(cross(va, vb) * cross(va, vc), 0.0);
}

bool intersectLP(const Line &l, const Point &p){
	Point va = l.p1 - p, vb = l.p2 - p;
	return EQ(cross(va, vb), 0.0);
}

bool intersectSS(const Segment &s, const Segment &t){
	Point a = s.p1, b = s.p2, c = t.p1, d = t.p2;
	return ccw(a, b, c) * ccw(a, b, d) <= 0
		&& ccw(c, d, a) * ccw(c, d, b) <= 0;
}

bool intersectSP(const Segment &s, const Point &p){
	return ccw(s.p1, s.p2, p) == ON_SEGMENT;
}

void solve()
{
	int n;
	cin >> n;
	Polygon g(n);
	for (int i = 0; i < n; i++){
		cin >> g[i];
	}
	
	bool res = true;
	
	if (n & 1) res = false;
	
	for (int i = 0; i < n / 2; i++){
		Segment s = {g[i], g[i + 1]};
		Segment t = {g[i + n / 2], (g[(i + n / 2 + 1) % g.size()])};
		if (!EQ(abs(s.p1 - s.p2), abs(t.p1 - t.p2))) res = false;
		if (!parallel(s, t)) res = false;
	}
	
	if (res){
		cout << (g[0] + g[n / 2]) / 2.0 << endl;
	}
	else {
		puts("NA");
	}
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout << fixed << setprecision(20);
	solve();
}