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>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
#define pb push_back
int dy[]={0, 0, 1, -1, 1, 1, -1, -1};
int dx[]={1, -1, 0, 0, 1, -1, -1, 1};

#define FOR(i,a,b) for (int i=(a);i<(b);i++)
#define RFOR(i,a,b) for (int i=(b)-1;i>=(a);i--)
#define REP(i,n) for (int i=0;i<(n);i++)
#define RREP(i,n) for (int i=(n)-1;i>=0;i--)
#define mp make_pair
#define fi first
#define sc second

typedef double R;//double long double の切り替え cmathの関数はオーバーロードに対応しているので問題ない
typedef complex<R> Point;
typedef pair<Point , Point> Line;
typedef pair<Point ,R > Circle;
typedef vector<Point> Poly;

#define EPS (1e-10)//誤差
#define EQ(a,b) (abs((a)-(b)) < EPS)//２つの実数が等しいか
#define EQV(a,b) ( EQ((a).real(), (b).real()) && EQ((a).imag(), (b).imag()) )//２つのベクトルが等しいか
#define ft first
#define sd second
#define pb push_back
 
#define FOR(i,a,b) for (int i=(a);i<(b);i++)
#define RFOR(i,a,b) for (int i=(b)-1;i>=(a);i--)
#define REP(i,n) for (int i=0;i<(n);i++)
#define RREP(i,n) for (int i=(n)-1;i>=0;i--)

R dot(Point a,Point b){//内積ok
    return (a.real() * b.real() + a.imag() * b.imag());
}
R cross(Point a,Point b){//外積ok
    return (a.real() * b.imag() - a.imag() * b.real());
}




bool is_orthogonal(Line a,Line b){//2直線の直行判定ok
    return EQ(dot(a.ft - a.sd,b.ft - b.sd),0.0);
}
bool is_parallel(Line a,Line b){//2直線の並行判定ok
    return EQ(cross(a.ft - a.sd,b.ft - b.sd),0.0);
}



////////////////////交差判定
int ccw(Point a,Point b,Point c){//ok
    b -= a; c -= a;
    if(cross(b,c) > EPS) return 1;//a→bで反時計周りに折れてb→c
    if(cross(b,c) < -EPS) return -1;//a→bで時計周りに折れてb→c
    if(dot(b,c) < -EPS) return 2;//c--a--b on same line
    if(norm(c) - norm(b) > EPS) return -2;//a--b--c(absじゃなくて二乗するのは差が出やすいから?)
    return 0;//a--c--bまたはb==c
}

bool is_intersection_ll(Line l,Line m){//２つの直線が交わるかok
    return abs(cross(l.sd - l.ft,m.sd - m.ft)) > EPS || //平行でない
        abs(cross(l.sd - l.ft,m.ft - l.ft)) < EPS; //平行だが同じ線
}

bool is_intersection_ls(Line l,Line s){//直線lと線分sが交わるか
    return cross(l.sd - l.ft, s.ft-l.ft)*       // s[0] is left of l
        cross(l.sd - l.ft, s.sd - l.ft) < EPS; // s[1] is right of l
}

bool is_intersection_lp(Line l,Point p){//直線lと点pが交わるか
    return abs(cross(l.sd - p,l.ft - p));
}

bool is_intersection_ss(Line a,Line b){//２つの線分が交わるかok
    return ccw(a.ft,a.sd,b.ft)*ccw(a.ft,a.sd,b.sd) <= 0 && ccw(b.ft,b.sd,a.ft)*ccw(b.ft,b.sd,a.sd) <= 0;
}

bool is_intersection_sp(Line s,Point p){//線分と点の交差判定 三角不等式の利用
    return abs(s.ft - p) + abs(s.sd - p) - abs(s.ft - s.sd) < EPS;
}


bool intersection_cc(Circle c1,Circle c2){//２つの円の交差判定ok
    return abs(c1.ft - c2.ft) - (c1.sd + c2.sd) < -EPS;
}


/////////////距離
R dis_lp(Line l,Point p){//直線lと点pの距離ok
    return abs(cross(l.sd - l.ft,p - l.ft)) / abs(l.sd - l.ft);
}

R dis_ll(Line l,Line m){//２つの直線の距離
    return is_intersection_ll(l,m) ? 0.0 : dis_lp(l,m.ft);
}

R dis_ls(Line l,Line s){//直線lと線分sの距離
    if(is_intersection_ls(l,s)) return 0.0;
    return min(dis_lp(l,s.ft),dis_lp(l,s.sd));
}

R dis_sp(Line s,Point p){//線分sと点pの距離ok
    if(dot(s.sd - s.ft,p - s.ft) < EPS) return abs(p - s.ft);
    if(dot(s.ft - s.sd,p - s.sd) < EPS) return abs(p - s.sd);
    return dis_lp(s,p);
}

R dis_ss(Line s,Line t){//２つの線分の距離ok
    if(is_intersection_ss(s,t)) return 0.0;
    return min(min(dis_sp(s,t.ft),dis_sp(s,t.sd)),
            min(dis_sp(t,s.ft),dis_sp(t,s.sd)));
}



//////////////射影と反射
Point projection(Line l,Point p){//射影を求めるok
    R t = dot(p - l.ft,l.ft - l.sd) / norm(l.ft - l.sd);
    return l.ft + t * (l.ft - l.sd);
}

Point reflection(Line l,Point p){//反射を求めるok
    return p + (R)2.0 * (projection(l,p) - p);
}




//////////////交点(交差する保証してないときは交差判定してからつかってね)

Point intersection_ll(Line l,Line m){//交差判定してるなら線分にも使えるok
    R A = cross(l.sd - l.ft,m.sd - m.ft);
    R B = cross(l.sd - l.ft,l.sd - m.ft);
    if(abs(A) < EPS && abs(B) < EPS) return m.ft;//同じ線
    //if(abs(A) < EPS)assert(false);//並行で交点なし
    return m.ft + B / A * (m.sd - m.ft);
}

Line intersection_of_two_circles(Circle c1,Circle c2){//ok ２つの円の交点をLineに入れて返す(r1 + r2 > sqrt((x1 - x2) ^ 2 + (y1 - y2) ^ 2))を満たす必要があるok
    R a =  abs(c2.ft - c1.ft);
    R b = c1.sd;
    R c = c2.sd;

    R rc = (a  * a + b * b - c * c) / (2.0 * a);
    R rs = sqrt(b * b - rc * rc);//C++ ではオーバーロードが可能であるため、sqrt または float 型を受け取る long double のオーバーロードを呼び出すことができます。 C プログラムでは、sqrt は常に double を受け取って返します。
    Point diff = (c2.ft - c1.ft) / a;

    Line p ;
    p.ft = c1.ft + diff * rc + diff * Point(0,1) * rs;
    p.sd = c1.ft + diff * rc + diff * Point(0,-1) * rs;

    return p;
}


/////////////////////////polygon

#define currP(P,i) P[(i) % P.size()]//今の頂点
#define nextP(P,i) P[((i) + 1)%P.size()]//次の頂点

int is_contains_p_in_Poly(Poly po,Point p){//点が多角形の内部(1)、境界(-1)、外部(0)のどこにあるかを判定ok
    bool in = false;
    REP(i,po.size()){
        Point a = currP(po,i) - p,b = nextP(po,i) - p;
        if(a.imag() > b.imag())swap(a,b);
        if(a.imag() < EPS && EPS < b.imag())
            if(cross(a,b) < -EPS) in = !in;
        if(EQ(cross(a,b),0.0) && dot(a,b) < EPS)return  -1;
    }
    return in;
}

R area2(Poly po){//多角形の面積の二倍を求めるok
    R A = 0.0;
    REP(i,po.size())
        A += cross(currP(po,i),nextP(po, i));
    return A;
}




///////////////////////////凸

bool comp_complex_real(Point a,Point b){//x→yの辞書順ok
    if(EQ(a.real(),b.real()))
        return b.imag() - a.imag() > EPS;
    return b.real() - a.real() > EPS;
}

Poly convex_hull(Poly ps){//凸包ok
    int n = ps.size(),k = 0;
    sort(ps.begin(),ps.end(),comp_complex_real);
    Poly ch(2*n);
    for(int i = 0;i < n;ch[k++] = ps[i++])// lower-hull
        while(k >= 2 && ccw(ch[k - 2],ch[k - 1],ps[i]) <= 0 && ccw(ch[k - 2],ch[k - 1],ps[i]) > -2) --k;//３つ目の条件は180度を含むときのみ必要
    for(int i = n - 2,t = k + 1;i >= 0;ch[k++] = ps[i--])//upper-hull
        while(k >= t && ccw(ch[k - 2],ch[k - 1],ps[i]) <= 0 && ccw(ch[k - 2],ch[k - 1],ps[i]) > -2) --k;//上に同じ
    ch.resize(k - 1);
    return ch;
}
#define prevP(P, i) P[(i+P.size()-1) % P.size()]
bool isconvex(Poly P){//凸性判定時計回り反時計周りに対応
    bool cl = false,ccl = false;
    for(int i = 0;i < P.size();++i){
        int c = ccw(prevP(P,i),currP(P,i),nextP(P,i));
        if(c == -2)continue;//180度を含むときのみ　360度も含むときはc == -2 || c == 0
        if(c == 1)ccl = true;
        else if(c == 2)cl = true;
        else return false;
    }
    return !(cl && ccl);
}


ll n;

Point p[100];
Line l[100];

int main(){
	cin >> n;


	REP(i,n) {
		int x,y;
		cin >> x >> y;
		p[i] = Point(x,y);
	}
	if(n % 2 == 1) {
		cout << "NA" << endl;
		return 0;
	}

	REP(i,n) {
		l[i] = Line(p[(i+1)%n],p[i]);
	}

	bool f = true;
	REP(i,n/2) {
		if(!is_parallel(l[i],l[i+n/2])) {
			f = false;
		}
		if(abs(abs(l[i].ft - l[i].sc) - abs(l[i+n/2].ft-l[i+n/2].sc)) > EPS) {
			f = false;
		}
	}

	if(!f){
		cout << "NA" << endl;
		return 0;
	}

	
	Point ans = intersection_ll(Line(p[0],p[n/2]),Line(p[1],p[n/2+1]));
	printf("%.10f %.10f\n",ans.real(),ans.imag());

	return 0;
}