#include <bits/stdc++.h>
using namespace std;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<vector<long long>> v(n);
  for (long long i = 0; i < n; ++i) {
    long long a;
    for (long long j = 0; j < 5; j++) {
      cin >> a;
      v[i].push_back(a);
    }
  }
  vector<long long> ans;
  for (long long i = 0; i < n; ++i) {
    bool ok = 1;
    for (long long j = 0; j < n; ++j) {
      if (i != j)
        for (long long k = j + 1; k < n; ++k) {
          if (k != i) {
            long long prod = 0;
            for (long long m = 0; m < 5; m++) {
              long long nk = (v[k][m] - v[i][m]);
              long long nj = (v[j][m] - v[i][m]);
              prod += nk * nj;
            }
            if (prod > 0) {
              ok = 0;
              break;
            }
          }
        }
    }
    if (ok) ans.push_back(i);
  }
  cout << ans.size() << "\n";
  for (long long i : ans) cout << i + 1 << " ";
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<vector<long long>> v(n);
  for (long long i = 0; i < n; ++i) {
    long long a;
    for (long long j = 0; j < 5; j++) {
      cin >> a;
      v[i].push_back(a);
    }
  }
  vector<long long> ans;
  for (long long i = 0; i < n; ++i) {
    bool ok = 1;
    for (long long j = 0; j < n; ++j) {
      if (i != j)
        for (long long k = j + 1; k < n; ++k) {
          if (k != i) {
            long long prod = 0;
            for (long long m = 0; m < 5; m++) {
              long long nk = (v[k][m] - v[i][m]);
              long long nj = (v[j][m] - v[i][m]);
              prod += nk * nj;
            }
            if (prod > 0) {
              ok = 0;
              break;
            }
          }
        }
    }
    if (ok) ans.push_back(i);
  }
  cout << ans.size() << "\n";
  for (long long i : ans) cout << i + 1 << " ";
  return 0;
}

```