#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(228);
int main() {
  int n;
  cin >> n;
  if (n > 100) {
    cout << 0 << '\n';
    return 0;
  }
  vector<vector<int> > a;
  for (int i = 0; i < n; i++) {
    vector<int> e;
    int x;
    for (int it = 0; it < 5; it++) {
      scanf("%d", &x);
      e.push_back(x);
    }
    a.push_back(e);
  }
  double pi = acos(-1);
  vector<int> ans;
  for (int i = 0; i < n; i++) {
    bool bad = false;
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < n; k++) {
        if (i != j && i != k && j != k) {
          long long scal = 0;
          long long len_a = 0;
          long long len_b = 0;
          for (int it = 0; it < 5; it++) {
            scal += (a[j][it] - a[i][it]) * (long long)(a[k][it] - a[i][it]);
            len_a += (a[j][it] - a[i][it]) * (long long)(a[j][it] - a[i][it]);
            len_b += (a[k][it] - a[i][it]) * (long long)(a[k][it] - a[i][it]);
          }
          double l_a = sqrt(len_a);
          double l_b = sqrt(len_b);
          double oreo = scal / (l_a * l_b);
          double ang = acos(oreo) * 180.0 / pi;
          if (ang < 90) {
            bad = true;
          }
        }
      }
    }
    if (!bad) {
      ans.push_back(i);
    }
  }
  cout << ans.size() << '\n';
  for (int c : ans) cout << c + 1 << ' ';
  cout << '\n';
}

### Prompt

Create a solution in cpp for 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;
mt19937 rnd(228);
int main() {
  int n;
  cin >> n;
  if (n > 100) {
    cout << 0 << '\n';
    return 0;
  }
  vector<vector<int> > a;
  for (int i = 0; i < n; i++) {
    vector<int> e;
    int x;
    for (int it = 0; it < 5; it++) {
      scanf("%d", &x);
      e.push_back(x);
    }
    a.push_back(e);
  }
  double pi = acos(-1);
  vector<int> ans;
  for (int i = 0; i < n; i++) {
    bool bad = false;
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < n; k++) {
        if (i != j && i != k && j != k) {
          long long scal = 0;
          long long len_a = 0;
          long long len_b = 0;
          for (int it = 0; it < 5; it++) {
            scal += (a[j][it] - a[i][it]) * (long long)(a[k][it] - a[i][it]);
            len_a += (a[j][it] - a[i][it]) * (long long)(a[j][it] - a[i][it]);
            len_b += (a[k][it] - a[i][it]) * (long long)(a[k][it] - a[i][it]);
          }
          double l_a = sqrt(len_a);
          double l_b = sqrt(len_b);
          double oreo = scal / (l_a * l_b);
          double ang = acos(oreo) * 180.0 / pi;
          if (ang < 90) {
            bad = true;
          }
        }
      }
    }
    if (!bad) {
      ans.push_back(i);
    }
  }
  cout << ans.size() << '\n';
  for (int c : ans) cout << c + 1 << ' ';
  cout << '\n';
}

```