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:
//-------------
//| ConvexCut |
//| H28.03.12 |
//-------------

//---includes---//
#include<iostream>
#include<cmath>
#include<iomanip>
#include<vector>
#include<algorithm>
using namespace std;

//---defines---//
#define MAX_N 100000
#define EPS 1e-14
#define EPS2 1e-6
#define INF 1e14
#define INF2 1e12

//---ints---//
int Q;
vector<pair<long double, long double>>vec;
long double X[MAX_N], Y[MAX_N];
long double X1[MAX_N], X2[MAX_N], X3[MAX_N], X4[MAX_N];
long double Y1[MAX_N], Y2[MAX_N], Y3[MAX_N], Y4[MAX_N];

bool paralel(long double X1, long double Y1, long double X2, long double Y2, long double X3, long double Y3, long double X4, long double Y4) {
	long double xp[4], yp[4]; 
	xp[0] = X1 - X2; yp[0] = Y1 - Y2; 
	xp[1] = X3 - X4; yp[1] = Y3 - Y4;
	if (yp[0] == 0 && yp[1] == 0) { return true; }
	if (yp[0] == 0 && yp[1] != 0) { return false; }
	if (yp[0] != 0 && yp[1] == 0) { return false; }
	if ((xp[0] / yp[0]) == (xp[1] / yp[1])) { return true; }
	return false;
}
pair<long double, long double>kouten(long double X1, long double Y1, long double X2, long double Y2, long double X3, long double Y3, long double X4, long double Y4) {
	if (paralel(X1, Y1, X2, Y2, X3, Y3, X4, Y4) == true) { 
		return make_pair(INF, INF); 
	}
	if (fabs(X1 - X2) < EPS) { X1 -= EPS; }
	if (fabs(X3 - X4) < EPS) { X3 -= EPS; }
	if (fabs(Y1 - Y2) < EPS) { Y1 -= EPS; }
	if (fabs(Y3 - Y4) < EPS) { Y3 -= EPS; }
	long double XP[4], YP[4], ZP[4];

	XP[0] = X1 - X2; YP[0] = Y1 - Y2;
	XP[1] = X3 - X4; YP[1] = Y3 - Y4; 
	ZP[0] = YP[0] / XP[0]; 
	ZP[1] = YP[1] / XP[1];
	long double X5 = X1, Y5 = Y3 - (X3 - X1)*ZP[1];
	long double K = (Y5 - Y1) / (ZP[0] - ZP[1]); 
	long double L = K*ZP[0]; 
	return make_pair(X1 + K, Y1 + L);
}
void solve() {
	for (int i = 0; i < Q; i++) { 
		X1[i] = X[i]; Y1[i] = Y[i]; 
		X2[i] = X[(i + Q / 2) % Q]; 
		Y2[i] = Y[(i + Q / 2) % Q]; 
		vec.push_back(kouten(X[0], Y[0], X[Q / 2], Y[Q / 2], X1[i], Y1[i], X2[i], Y2[i])); 
	}
	sort(vec.begin(), vec.end()); 
	for (int i = 0; i < vec.size() - 1; i++) { 
		long double X_1 = vec[i].first, X_2 = vec[i + 1].first;
		long double Y_1 = vec[i].second, Y_2 = vec[i + 1].second;
		if ((X_1 < INF2 || Y_1 < INF2) && (X_2 < INF2 || Y_2 < INF2)) {
			if (fabs(X_1 - X_2) > EPS2 || fabs(Y_1 - Y_2) > EPS2) {
				cout << "NA" << endl;
				return;
			}
		}
	}
	cout << fixed << setprecision(15) << vec[0].first << ' ';
	cout << fixed << setprecision(15) << vec[0].second << endl;
}
int main() { 
	cin >> Q; 
	for (int i = 0; i < Q; i++) { 
		cin >> X[i] >> Y[i]; 
	}
	solve();
	return 0;
}