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>
#define rep(i,a,...) for(int i = (a)*(strlen(#__VA_ARGS__)!=0);i<(int)(strlen(#__VA_ARGS__)?__VA_ARGS__:(a));++i)
#define per(i,a,...) for(int i = (strlen(#__VA_ARGS__)?__VA_ARGS__:(a))-1;i>=(int)(strlen(#__VA_ARGS__)?(a):0);--i)
#define foreach(i, n) for(auto &i:(n))
#define pii pair<int, int>
#define pll pair<long long, long long>
#define all(x) (x).begin(), (x).end()
#define bit(x) (1ll << (x))
using ll = long long;
//const ll MOD = (ll)1e9+7;
const ll MOD = 998244353;
const int INF = (ll)1e9+7;
const ll INFLL = (ll)1e18;
using namespace std;
template<class t>
using vvector = vector<vector<t>>;
template<class t>
using vvvector = vector<vector<vector<t>>>;
template<class t>
using priority_queuer = priority_queue<t, vector<t>, greater<t>>;
template<class t, class u> bool chmax(t &a, u b){if(a<b){a=b;return true;}return false;}
template<class t, class u> bool chmin(t &a, u b){if(a>b){a=b;return true;}return false;}
#ifdef DEBUG
#define debug(x) cout<<"LINE "<<__LINE__<<": "<<#x<<" = "<<x<<endl;
#else
#define debug(x) (void)0
#endif

namespace templates{
  ll modpow(ll x, ll b){
    ll res = 1;
    while(b){
      if(b&1)res = res * x % MOD;
      x = x * x % MOD;
      b>>=1;
    }
    return res;
  }

  ll modinv(ll x){
    return modpow(x, MOD-2);
  }

  bool was_output = false;
  template<class t>
  void output(t a){
    if(was_output)cout << " ";
    cout << a;
    was_output = true;
  }
  void outendl(){
    was_output = false;
    cout << endl;
  }
  ll in(){
    ll res;
    scanf("%lld", &res);
    return res;
  }

  template<class t>
  istream& operator>>(istream&is, vector<t>&x){
    for(auto &i:x)is >> i;
    return is;
  }

  template<class t, class u>
  istream& operator>>(istream&is, pair<t, u>&x){
    is >> x.first >> x.second;
    return is;
  }

  template<class t>
  void in(t&x){
    cin >> x;
  }

  template<class t>
  t in(){
    t res; cin >> res; return res;
  }

  template<class t>
  void out(t x){
    cout << x;
  }

  template<class t>
  vector<t> sorted(vector<t> line,function<bool(t,t)> comp=[](t a,t b){return a<b;}){
    sort(line.begin(),line.end(),comp);
    return line;
  }

  template<class t>
  vector<t> reversed(vector<t> line){
    reverse(line.begin(),line.end());
    return line;
  }
}

using namespace templates;

using P = pair<double,double>;
P merge(P a,P b){
  return P((a.first+b.first)/2,(a.second+b.second)/2);
}

bool check(P a,P b){
  return max(abs(a.first-b.first),abs(a.second-b.second)) < 0.00001;
}

int main(){
  int n = in();
  if(n&1){
    cout << "NA" << endl;
    return 0;
  }
  vector<P> line(n);
  foreach(i,line)cin>>i;
  vector<P> mids;
  rep(i,n/2){
    mids.emplace_back(merge(line[i],line[i+n/2]));
  }
  rep(i,n/2){
    rep(j,n/2){
      if(!check(mids[i],mids[j])){
        cout << "NA" << endl;
        return 0;
      }
    }
  }
  printf("%.20lf %.20lf\n",mids[0].first,mids[0].second);
  return 0;
}