#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<tuple<long long, long long, long long, long long, long long>> vet1(n);
  vector<long long> res;
  for (int i = 0; i < n; ++i) {
    long long a, b, c, d, e;
    cin >> a >> b >> c >> d >> e;
    auto aux = make_tuple(a, b, c, d, e);
    vet1[i] = aux;
  }
  for (int i = 0; i < n; ++i) {
    long long a0, b0, c0, d0, e0;
    bool flag1 = true;
    tie(a0, b0, c0, d0, e0) = vet1[i];
    for (int j = 0; j < n && flag1; ++j)
      if (j != i) {
        long long a1, b1, c1, d1, e1;
        tie(a1, b1, c1, d1, e1) = vet1[j];
        a1 -= a0;
        b1 -= b0;
        c1 -= c0;
        d1 -= d0;
        e1 -= e0;
        for (int k = 0; k < n && flag1; ++k)
          if (k != j && k != i) {
            long long a2, b2, c2, d2, e2;
            tie(a2, b2, c2, d2, e2) = vet1[k];
            a2 -= a0;
            b2 -= b0;
            c2 -= c0;
            d2 -= d0;
            e2 -= e0;
            long long val = a1 * a2 + b1 * b2 + c1 * c2 + d1 * d2 + e1 * e2;
            if (val > 0) {
              flag1 = false;
            }
          }
      }
    if (flag1) {
      res.push_back(i + 1);
    }
  }
  cout << res.size() << endl;
  for (int i = 0; i < res.size(); ++i) {
    cout << res[i] << " \n"[i == res.size() - 1];
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<tuple<long long, long long, long long, long long, long long>> vet1(n);
  vector<long long> res;
  for (int i = 0; i < n; ++i) {
    long long a, b, c, d, e;
    cin >> a >> b >> c >> d >> e;
    auto aux = make_tuple(a, b, c, d, e);
    vet1[i] = aux;
  }
  for (int i = 0; i < n; ++i) {
    long long a0, b0, c0, d0, e0;
    bool flag1 = true;
    tie(a0, b0, c0, d0, e0) = vet1[i];
    for (int j = 0; j < n && flag1; ++j)
      if (j != i) {
        long long a1, b1, c1, d1, e1;
        tie(a1, b1, c1, d1, e1) = vet1[j];
        a1 -= a0;
        b1 -= b0;
        c1 -= c0;
        d1 -= d0;
        e1 -= e0;
        for (int k = 0; k < n && flag1; ++k)
          if (k != j && k != i) {
            long long a2, b2, c2, d2, e2;
            tie(a2, b2, c2, d2, e2) = vet1[k];
            a2 -= a0;
            b2 -= b0;
            c2 -= c0;
            d2 -= d0;
            e2 -= e0;
            long long val = a1 * a2 + b1 * b2 + c1 * c2 + d1 * d2 + e1 * e2;
            if (val > 0) {
              flag1 = false;
            }
          }
      }
    if (flag1) {
      res.push_back(i + 1);
    }
  }
  cout << res.size() << endl;
  for (int i = 0; i < res.size(); ++i) {
    cout << res[i] << " \n"[i == res.size() - 1];
  }
  return 0;
}

```