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:
#define __USE_MINGW_ANSI_STDIO 0
#include <bits/stdc++.h>

using namespace std;
using ll = long long;
#define int ll
using VI = vector<int>;
using VVI = vector<VI>;
using PII = pair<int, int>;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define PB push_back

const ll LLINF = (1LL<<60);
const int INF = (1LL<<30);
const int MOD = 1000000007;

template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }
template <typename T> bool IN(T a, T b, T x) { return a<=x&&x<b; }
template<typename T> T ceil(T a, T b) { return a/b + !!(a%b); }
template<class S,class T>
ostream &operator <<(ostream& out,const pair<S,T>& a){
  out<<'('<<a.first<<','<<a.second<<')';
  return out;
}
template<class T>
ostream &operator <<(ostream& out,const vector<T>& a){
  out<<'[';
  REP(i, a.size()) {out<<a[i];if(i!=a.size()-1)out<<',';}
  out<<']';
  return out;
}

int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};

const double EPS = 1e-8;

using R = long double;
using P = complex<R>;
using L = pair<P,P>;
using G = vector<P>;
struct C {
  P c; R r;
  C() {}
  C(const P &a, const R &b) : c(a), r(b) {}
};
struct S : public L {
  S() {}
  S(const P &a, const P &b) : L(a,b) {}
};

inline int sgn(const R& r) { return (r>EPS) - (r<-EPS); }
inline R dot(const P& a, const P& b) {
  return real(a)*real(b) + imag(a)*imag(b);
}
inline R det(const P& a, const P& b) {
  return real(a)*imag(b) - imag(a)*real(b);
}
inline P vec(const L& l) {return l.second - l.first;}
namespace std {
	bool operator < (const P& a, const P& b) {
		return sgn(real(a-b)) ? real(a-b) < 0 : sgn(imag(a-b)) < 0;
	}
	bool operator == (const P& a, const P& b) {
		return sgn(real(a-b)) == 0 && sgn(imag(a-b)) == 0;
	}
  bool cmp_y (const P& a, const P& b) {
    return sgn(imag(a-b)) ? imag(a-b) < 0 : sgn(real(a-b)) < 0;
  }
}

// P,L,Sについて入力
inline istream& operator>>(istream& is, P& p) {
  R x, y;
  is >> x >> y;
  p = P(x, y);
  return is;
}
inline istream& operator>>(istream& is, L& l) {
  P a, b;
  is >> a >> b;
  l = L(a, b);
  return is;
}
inline istream& operator>>(istream& is, S& s) {
  P a, b;
  is >> a >> b;
  s = S(a, b);
  return is;
}

inline bool parallel(L a, L b) {return sgn(det(vec(a), vec(b))) == 0;}

P p[55];
signed main(void)
{
  cin.tie(0);
  ios::sync_with_stdio(false);

  int n;
  cin >> n;
  REP(i, n) cin >> p[i];

  if(n%2) {
    cout << "NA" << endl;
    return 0;
  }

  bool flag = true;
  REP(i, n) {
    // p[i]-p[i+1] と p[i+n/2]-p[i+n/2+1] について
    L l1(p[i], p[(i+1)%n]), l2(p[(i+n/2)%n], p[(i+n/2+1)%n]);
    if(!parallel(l1, l2) || sgn(abs(vec(l1))-abs(vec(l2))) != 0) {
      flag = false;
    }
  }

  cout << fixed << setprecision(9) << endl;
  if(flag) cout << (p[0].real()+p[n/2].real())/2.0 << " " << (p[0].imag()+p[n/2].imag())/2.0 << endl;
  else cout << "NA" << endl;

  return 0;
}