#include <bits/stdc++.h>
using namespace std;
struct pt {
  long double o, t, th, f, fv;
  pt(long double a, long double b, long double c, long double d,
     long double e) {
    o = a;
    t = b;
    th = c;
    f = d;
    fv = e;
  }
  pt() {}
  pt operator-(const pt &a) const {
    pt b;
    b.o = o - a.o;
    b.t = t - a.t;
    b.th = th - a.th;
    b.f = f - a.f;
    b.fv = fv - a.fv;
    return b;
  }
};
long double dot(pt a, pt b) {
  return a.o * b.o + a.t * b.t + a.th * b.th + a.f * b.f + a.fv * b.fv;
}
int main() {
  int n;
  cin >> n;
  vector<pt> v(n);
  long double o, t, th, f, fv;
  vector<int> ase;
  for (int i = 0; i < n; i++) {
    cin >> o >> t >> th >> f >> fv;
    v[i] = pt(o, t, th, f, fv);
  }
  for (int i = 0; i < n; i++) {
    int uff = 0;
    for (int j = 0; j < n; j++) {
      if (j == i) continue;
      for (int k = j + 1; k < n; k++) {
        if (k == i) continue;
        if (acos(dot(v[j] - v[i], v[k] - v[i]) /
                 (sqrt(dot(v[j] - v[i], v[j] - v[i])) *
                  sqrt(dot(v[k] - v[i], v[k] - v[i])))) < acos(-1) / 2) {
          uff = 1;
          break;
        }
      }
      if (uff) break;
    }
    if (!uff) ase.push_back(i);
  }
  cout << ase.size() << endl;
  for (int i = 0; i < ase.size(); i++) cout << ase[i] + 1 << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
struct pt {
  long double o, t, th, f, fv;
  pt(long double a, long double b, long double c, long double d,
     long double e) {
    o = a;
    t = b;
    th = c;
    f = d;
    fv = e;
  }
  pt() {}
  pt operator-(const pt &a) const {
    pt b;
    b.o = o - a.o;
    b.t = t - a.t;
    b.th = th - a.th;
    b.f = f - a.f;
    b.fv = fv - a.fv;
    return b;
  }
};
long double dot(pt a, pt b) {
  return a.o * b.o + a.t * b.t + a.th * b.th + a.f * b.f + a.fv * b.fv;
}
int main() {
  int n;
  cin >> n;
  vector<pt> v(n);
  long double o, t, th, f, fv;
  vector<int> ase;
  for (int i = 0; i < n; i++) {
    cin >> o >> t >> th >> f >> fv;
    v[i] = pt(o, t, th, f, fv);
  }
  for (int i = 0; i < n; i++) {
    int uff = 0;
    for (int j = 0; j < n; j++) {
      if (j == i) continue;
      for (int k = j + 1; k < n; k++) {
        if (k == i) continue;
        if (acos(dot(v[j] - v[i], v[k] - v[i]) /
                 (sqrt(dot(v[j] - v[i], v[j] - v[i])) *
                  sqrt(dot(v[k] - v[i], v[k] - v[i])))) < acos(-1) / 2) {
          uff = 1;
          break;
        }
      }
      if (uff) break;
    }
    if (!uff) ase.push_back(i);
  }
  cout << ase.size() << endl;
  for (int i = 0; i < ase.size(); i++) cout << ase[i] + 1 << endl;
  return 0;
}

```