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 <bits/stdc++.h>
using namespace std;

typedef double frac;
const double EPS=1e-10;
double add(double l, double r){
    if(abs(l+r)<EPS*(abs(l)+abs(r))) return 0;
    return l+r;
}
typedef double frac;
struct Pt{
    frac x, y;
    Pt(frac xx=0, frac yy=0){x=xx; y=yy;}
    Pt& operator=(const Pt &rhs){
        x=rhs.x; y=rhs.y; return *this;
    }
    Pt& operator+=(const Pt &rhs){
        x = add(x,rhs.x); y = add(y,rhs.y); return *this;
    }
    Pt& operator-=(const Pt &rhs){
        x = add(x,-rhs.x); y = add(y,-rhs.y); return *this;
    }
    Pt& operator*=(const frac &rhs){
        x *= rhs; y *= rhs; return *this;
    }
    Pt& operator/=(const frac &rhs){
        x /= rhs; y /= rhs; return *this;
    }
};
Pt operator+(const Pt &lhs, const Pt &rhs){
    Pt sum=lhs; sum += rhs; return sum;
}
Pt operator-(const Pt &lhs, const Pt &rhs){
    Pt sum=lhs; sum -= rhs; return sum;
}
Pt operator*(const Pt &lhs, const frac &rhs){
    Pt sum=lhs; sum *= rhs; return sum;
}
Pt operator*(const frac &lhs, const Pt &rhs){
    Pt sum=rhs; sum *= lhs; return sum;
}
Pt operator/(const Pt &lhs, const frac &rhs){
    Pt sum=lhs; sum /= rhs; return sum;
}
bool operator==(const Pt &lhs, const Pt &rhs){return lhs.x==rhs.x && lhs.y==rhs.y;}
bool operator!=(const Pt &lhs, const Pt &rhs){return !(lhs==rhs);}
bool operator<(const Pt &lhs, const Pt &rhs){return lhs.x!=rhs.x ? lhs.x<rhs.x : lhs.y<rhs.y;}
bool operator>(const Pt &lhs, const Pt &rhs){return rhs<lhs;}
bool operator<=(const Pt &lhs, const Pt &rhs){return !(lhs>rhs);}
bool operator>=(const Pt &lhs, const Pt &rhs){return !(lhs<rhs);}
bool operator<(Pt lhs, Pt rhs){
    return lhs.x!=rhs.x ? lhs.x<rhs.x : lhs.y<rhs.y;
}
istream& operator>>(istream &is, Pt &rhs){
    double x, y;
    is >> x >> y;
    rhs=Pt(x,y);
    return is;
}
ostream& operator<<(ostream &os, Pt rhs){
    os << rhs.x << " " << rhs.y;
    return os;
}
frac dot(Pt a, Pt b){return a.x*b.x+a.y*b.y;}//内積
frac cross(Pt a, Pt b){return a.x*b.y-a.y*b.x;}//外積
//qが線分p1-p2上にあればtrue
bool on_seg(Pt p1, Pt p2, Pt q){
    return cross(p1-q,p2-q)==0 && dot(p1-q,p2-q)<=0;
}
//intersection　交点を求める
Pt insec(Pt p1, Pt p2, Pt q1, Pt q2){
    return p1+(cross(q2-q1,q1-p1)/cross(q2-q1,p2-p1))*(p2-p1);
}
int main(){
	int n;  cin >> n;
    vector<Pt> p(n+1);
    for(int i=0; i<n; i++) cin >> p[i];
    p[n]=p[0];

    bool flg=1;
    if(n%2==1){
        flg=0;
    }else{
        for(int i=0, j=n/2; i<n/2; ++i,++j){
            if(p[i]-p[i+1]!=-1*(p[j]-p[j+1])){
                flg=0;
                break;
            }
        }
    }

    if(flg){
        cout << fixed << setprecision(5);
        cout << (p[0]+p[n/2])/2 << endl;
    }else{
        cout << "NA" << endl;
    }
}
