#include <bits/stdc++.h>
using namespace std;
template <typename T>
istream &operator>>(istream &in, vector<T> &a) {
  for (auto &x : a) {
    in >> x;
  }
  return in;
}
template <typename T>
ostream &operator<<(ostream &out, const vector<T> &a) {
  for (auto x : a) {
    out << x << " ";
  }
  return out;
}
struct Point {
  long long x1, x2, x3, x4, x5;
  Point() {}
  void in() { scanf("%I64d %I64d %I64d %I64d %I64d", &x1, &x2, &x3, &x4, &x5); }
};
struct Vector {
  long long x1, x2, x3, x4, x5;
  long double len;
  Vector(const Point &a, const Point &b) {
    x1 = b.x1 - a.x1;
    x2 = b.x2 - a.x2;
    x3 = b.x3 - a.x3;
    x4 = b.x4 - a.x4;
    x5 = b.x5 - a.x5;
    len = sqrtl(x1 * x1 + x2 * x2 + x3 * x3 + x4 * x4 + x5 * x5);
  }
};
long double check(const Vector &a, const Vector &b) {
  long double up =
      a.x1 * b.x1 + a.x2 * b.x2 + a.x3 * b.x3 + a.x4 * b.x4 + a.x5 * b.x5;
  return acos(up / (a.len * b.len)) * 180 / acos(-1);
  ;
}
const int maxn = 1e3 + 7;
Point vec[maxn];
void solve() {
  int n;
  scanf("%d", &n);
  vector<int> pos;
  for (int i = 0; i < n; ++i) {
    vec[i].in();
  }
  bool bad = true;
  for (int i = 0; i < n; ++i) {
    bad = true;
    for (int j = 0; j < n && bad; ++j) {
      for (int k = 0; k < n && bad; ++k) {
        if (i == j || j == k || i == k) {
          continue;
        }
        if (check(Vector(vec[i], vec[j]), Vector(vec[i], vec[k])) < 90) {
          bad = false;
        }
      }
    }
    if (bad) {
      pos.push_back(i + 1);
    }
  }
  printf("%d\n", pos.size());
  for (int i = 0; i < pos.size(); ++i) {
    printf("%d\n", pos[i]);
  }
}
int main() {
  solve();
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

You are given set of n points in 5-dimensional space. The points are labeled from 1 to n. No two points coincide.

We will call point a bad if there are different points b and c, not equal to a, from the given set such that angle between vectors <image> and <image> is acute (i.e. strictly less than <image>). Otherwise, the point is called good.

The angle between vectors <image> and <image> in 5-dimensional space is defined as <image>, where <image> is the scalar product and <image> is length of <image>.

Given the list of points, print the indices of the good points in ascending order.

Input

The first line of input contains a single integer n (1 ≤ n ≤ 103) — the number of points.

The next n lines of input contain five integers ai, bi, ci, di, ei (|ai|, |bi|, |ci|, |di|, |ei| ≤ 103) — the coordinates of the i-th point. All points are distinct.

Output

First, print a single integer k — the number of good points.

Then, print k integers, each on their own line — the indices of the good points in ascending order.

Examples

Input

6
0 0 0 0 0
1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1


Output

1
1


Input

3
0 0 1 2 0
0 0 9 2 0
0 0 5 9 0


Output

0

Note

In the first sample, the first point forms exactly a <image> angle with all other pairs of points, so it is good.

In the second sample, along the cd plane, we can see the points look as follows:

<image>

We can see that all angles here are acute, so no points are good.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T>
istream &operator>>(istream &in, vector<T> &a) {
  for (auto &x : a) {
    in >> x;
  }
  return in;
}
template <typename T>
ostream &operator<<(ostream &out, const vector<T> &a) {
  for (auto x : a) {
    out << x << " ";
  }
  return out;
}
struct Point {
  long long x1, x2, x3, x4, x5;
  Point() {}
  void in() { scanf("%I64d %I64d %I64d %I64d %I64d", &x1, &x2, &x3, &x4, &x5); }
};
struct Vector {
  long long x1, x2, x3, x4, x5;
  long double len;
  Vector(const Point &a, const Point &b) {
    x1 = b.x1 - a.x1;
    x2 = b.x2 - a.x2;
    x3 = b.x3 - a.x3;
    x4 = b.x4 - a.x4;
    x5 = b.x5 - a.x5;
    len = sqrtl(x1 * x1 + x2 * x2 + x3 * x3 + x4 * x4 + x5 * x5);
  }
};
long double check(const Vector &a, const Vector &b) {
  long double up =
      a.x1 * b.x1 + a.x2 * b.x2 + a.x3 * b.x3 + a.x4 * b.x4 + a.x5 * b.x5;
  return acos(up / (a.len * b.len)) * 180 / acos(-1);
  ;
}
const int maxn = 1e3 + 7;
Point vec[maxn];
void solve() {
  int n;
  scanf("%d", &n);
  vector<int> pos;
  for (int i = 0; i < n; ++i) {
    vec[i].in();
  }
  bool bad = true;
  for (int i = 0; i < n; ++i) {
    bad = true;
    for (int j = 0; j < n && bad; ++j) {
      for (int k = 0; k < n && bad; ++k) {
        if (i == j || j == k || i == k) {
          continue;
        }
        if (check(Vector(vec[i], vec[j]), Vector(vec[i], vec[k])) < 90) {
          bad = false;
        }
      }
    }
    if (bad) {
      pos.push_back(i + 1);
    }
  }
  printf("%d\n", pos.size());
  for (int i = 0; i < pos.size(); ++i) {
    printf("%d\n", pos[i]);
  }
}
int main() {
  solve();
  return 0;
}

```