Problem: The Galaxy contains n planets, there are many different living creatures inhabiting each planet. And each creature can get into troubles! Space rescuers know it perfectly well and they are always ready to help anyone who really needs help. All you need to do is call for them. 

Now the space rescuers plan to build the largest in the history of the Galaxy rescue station; however, the rescue station's location is yet to be determined. As some cases are real emergencies, the rescuers want to find such a point in the Galaxy from which it would be possible to get to the remotest planet in the minimum possible time. In other words, the rescuers need such point in the space that the distance between it and the planet remotest from it was minimal (if we compare this point with all other possible points in the space). Unfortunately, the rescuers can't sole this problem.

As the planets are quite remote from each other, they can be considered as points in Euclidean three-dimensional space. The distance between points (xi, yi, zi) and (xj, yj, zj) can be calculated by the formula <image>. The rescue station can be positioned in any point in the space. It can also coincide with some planet. 

Galaxy is in danger! Save the space rescuers and find the required point for them.

Input

The first line of the input file contains integer n — the number of planets (1 ≤ N ≤ 100). Each of the following n lines contains information about the planets. The i-th line contains three integers xi, yi, zi — the coordinates of the i-th planet ( - 104 ≤ xi, yi, zi ≤ 104, 1 ≤ i ≤ n). No two planets coincide.

Output

Print on the first line of the output file three space-separated real numbers x0, y0, z0 — the coordinates for the future base. If there are several solutions, you are allowed to print any of them. The answer will be accepted if the distance from this point to the remotest planet will differ from the juries' variant in no more than 10 - 6 in absolute or relative value.

Examples

Input

5
5 0 0
-5 0 0
0 3 4
4 -3 0
2 2 -2


Output

0.000 0.000 0.000

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
using namespace std;
template <class T>
T sqr(T x) {
  return x * x;
}
template <class T>
T isSquare(T x) {
  int y = sqrt(x + 0.5);
  return (y * y) == x;
}
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <class T1, class T2>
T1 chmin(T1& x, const T2& y) {
  if (T1(y) < x) return x = y;
  return x;
}
template <class T1, class T2>
T1 chmax(T1& x, const T2& y) {
  if (T1(y) > x) return x = y;
  return x;
}
template <class T1, class T2>
ostream& operator<<(ostream& os, const pair<T1, T2>& p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << '{';
  bool was = false;
  for (const T& x : v) {
    if (was) os << ", ";
    was = true;
    os << x;
  }
  os << '}';
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const set<T>& v) {
  os << '[';
  bool was = false;
  for (const T& x : v) {
    if (was) os << ", ";
    was = true;
    os << x;
  }
  os << ']';
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const multiset<T>& v) {
  os << '[';
  bool was = false;
  for (const T& x : v) {
    if (was) os << ", ";
    was = true;
    os << x;
  }
  os << ']';
  return os;
}
template <class T1, class T2>
ostream& operator<<(ostream& os, const map<T1, T2>& m) {
  os << '<';
  bool was = false;
  for (const auto& x : m) {
    if (was) os << ", ";
    was = true;
    os << x;
  }
  os << '>';
  return os;
}
const double EPS = 1e-8;
const int INF = (int)2e9;
const long long LINF = (long long)2e18;
const int MOD = (int)1e9 + 7;
const int MAXN = (int)110;
const int N = 70;
struct KArs {
  double x, y, z;
  KArs() {}
  inline KArs(double x, double y, double z) : x(x), y(y), z(z) {}
  inline double dist(const KArs& other) const {
    return sqr(x - other.x) + sqr(y - other.y) + sqr(z - other.z);
  }
};
int n;
double ax[MAXN], ay[MAXN], az[MAXN];
int miX, miY, miZ, maX, maY, maZ;
double best = 2e15;
KArs pt;
inline double f(const double& x, const double& y, const double& z) {
  double res = 0;
  for (int i = 1; i <= n; ++i) {
    double dx = x - ax[i], dy = y - ay[i], dz = z - az[i];
    double r = dx * dx + dy * dy + dz * dz;
    if (res < r) res = r;
  }
  return res;
}
inline double fZ(double x, double y) {
  double l = miZ, r = maZ;
  for (int iter = 0; iter <= N; iter++) {
    double md1 = l + (r - l) / 3;
    double md2 = r - (r - l) / 3;
    if (f(x, y, md1) < f(x, y, md2)) {
      r = md2;
    } else {
      l = md1;
    }
  }
  pt.z = r;
  return f(x, y, r);
}
inline double fY(double x) {
  double l = miY, r = maY;
  for (int iter = 0; iter <= N; iter++) {
    double md1 = l + (r - l) / 3;
    double md2 = r - (r - l) / 3;
    if (fZ(x, md1) < fZ(x, md2)) {
      r = md2;
    } else {
      l = md1;
    }
  }
  pt.y = r;
  return fZ(x, r);
}
void fX() {
  double l = miX, r = maX;
  for (int iter = 0; iter <= N; iter++) {
    double md1 = l + (r - l) / 3;
    double md2 = r - (r - l) / 3;
    if (fY(md1) < fY(md2)) {
      r = md2;
    } else {
      l = md1;
    }
  }
  pt.x = r;
}
int main() {
  ios_base::sync_with_stdio(false);
  miX = miY = miZ = 1e4;
  maX = maY = maZ = -1e4;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> ax[i] >> ay[i] >> az[i];
    chmin(miX, ax[i]), chmax(maX, ax[i]);
    chmin(miY, ay[i]), chmax(maY, ay[i]);
    chmin(miZ, az[i]), chmax(maZ, az[i]);
  }
  fX();
  cout << fixed << setprecision(8) << pt.x;
  cout << ' ';
  cout << fixed << setprecision(8) << pt.y;
  cout << ' ';
  cout << fixed << setprecision(8) << pt.z;
}