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>

using namespace std;
typedef long long ll;
typedef pair<int,int> PI;
const double EPS=1e-3;
#define rep(i,n) for(int i=0;i<(int)(n);++i)
#define F first
#define S second
#define mp(a,b) make_pair(a,b)
#define pb(a) push_back(a)
#define SZ(a) (int)((a).size())
#define ALL(a) (a).begin(),(a).end()
#define RALL(a) a.rbegin(),a.rend()
#define FLL(a,b) memset((a),b,sizeof(a))
#define CLR(a) FLL(a,0)
#define declare(a,it) __typeof(a) it=(a)
#define FOR(it,a) for(declare(a.begin(),it);it!=a.end();++it)
#define FORR(it,a) for(declare(a.rbegin(),it);it!=a.rend();++it)
template<class T,class U> ostream& operator<< (ostream& o, const pair<T,U>& v){return o << "(" << v.F << ", " << v.S << ")";}
template<class T> ostream& operator<< (ostream& o, const vector<T>& v){o << "{";rep(i,SZ(v)) o << (i?", ":"") << v[i];return o << "}";}
int dx[]={0,1,0,-1,1,1,-1,-1};
int dy[]={1,0,-1,0,-1,1,1,-1};
int s2i(string& a){stringstream ss(a);int r;ss>>r;return r;}
int geti(){int n;scanf("%d", &n);return n;}

int x[100],y[100];



int main(int argc, char *argv[])
{
  int n;
  cin >> n;

  double sumx = 0;
  double sumy = 0;
  for(int i = 0; i < n; ++i){
    cin >> x[i] >> y[i];
    sumx += x[i];
    sumy += y[i];
  }
  sumx /= n;
  sumy /= n;
  
  bool ok = true;
  for(int i = 0; i < n; ++i){
    double ax = x[(i+1)%n] - x[i];
    double ay = y[(i+1)%n] - y[i];
    double bx = x[(i+1+n/2)%n] - x[(i+n/2)%n];
    double by = y[(i+1+n/2)%n] - y[(i+n/2)%n];
    ok &= abs(ax*by - bx*ay) < EPS;
    ok &= abs(ax*ax+ay*ay-(bx*bx + by*by)) < EPS;
    continue;
    ok &= abs((ax*bx+ay*by)/sqrt(ax*ax+ay*ay)/sqrt(bx*bx+by*by)-
              cos(acos(-1)*2/n)) < EPS;
  }
  
  if(ok && ~n%2) printf("%.10f %.10f\n", sumx, sumy);
  else cout << "NA" << endl;
  
  return 0;
}