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 <cstdio>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <complex>
#include <ctime>
#include <cstdlib>
using namespace std;

// * XY座標
#define X real()
#define Y imag()

// * 点の表現
typedef complex<double> P;

// * 許容する誤差ε
const double EPS = 1e-8;
// * Infinity
const double INF = 1e12;
// * 円周率
const double PI = acos(-1.0);

// * 2つのスカラーが等しいかどうか
bool equal(double a, double b){
	return ( fabs( a - b ) < EPS );
}

// * 内積 (dot product) : a・b = |a||b|cosΘ
double dot(P a, P b){
	return (a.real() * b.real() + a.imag() * b.imag());
}

// 外積 (cross product) : a×b = |a||b|sinΘ
double cross(P a, P b){
	return (a.real() * b.imag() - a.imag() * b.real());
}

// a1,a2を通る直線 と b1,b2を通る直線 が平行かどうか返す.
bool is_parallel(P a1, P a2, P b1, P b2){
	return equal( cross(a1-a2, b1-b2), 0.0 );
}

// a1,a2を通る直線 と b1,b2を通る直線の交点計算
P intersection(P a1, P a2, P b1, P b2){
	P a = a2 - a1;
	P b = b2 - b1;
	return a1 + a * cross(b, b1-a1) / cross(b, a);
}

int main(){
	int n;
	scanf("%d", &n);
	
	vector<P> vp(n);
	for(int i=0 ; i < n ; i++ ){
		double x, y;
		scanf("%lf %lf", &x, &y);
		vp[i] = P(x,y);
	}
	// used[i] := 辺 i について平行で長さが等しい辺が存在するかどうか
	bool used[51] = {false};
	for(int i=0 ; i < n/2 ; i++ ){
		if( used[i] ) continue;
		int j = i + n/2;
		// 辺の長さ
		double a = abs(vp[i] - vp[(i+1)%n]);
		double b = abs(vp[j] - vp[(j+1)%n]);
		bool flag = is_parallel( vp[i], vp[(i+1)%n], vp[j], vp[(j+1)%n]);
		if( flag && equal(a,b) ){
			used[i] = used[j] = true;
		}
	}
	
	int k = 0;
	for(int i=0 ; i < n ; i++ ){
		if( used[i] ){
			k++;
		}
	}
	// 頂点数が奇数 かつ すべての辺について平行で長さの等しい辺が存在するときは点Pが存在する.
	if( n%2 == 1 ){
		printf("NA\n");
	}else if( k == n ){
		P p = intersection(vp[0], vp[n/2], vp[1], vp[n/2+1]);
		printf("%.6f %.6f\n", p.X, p.Y);
	}else{
		printf("NA\n");
	}
}