#include <bits/stdc++.h>
using namespace std;
const int N = 1003;
int v[N][6];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 5; j++) cin >> v[i][j];
  if (n > 25) return puts("0"), 0;
  vector<int> ans;
  for (int i = 0; i < n; i++) {
    int cnt = 0;
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      for (int k = 0; k < n; k++) {
        if (k == i || k == j) continue;
        int cur = 0;
        for (int z = 0; z < 5; z++) {
          cur += (v[j][z] - v[i][z]) * (v[k][z] - v[i][z]);
        }
        cnt |= cur > 0;
      }
    }
    if (!cnt) ans.push_back(i + 1);
  }
  cout << ans.size() << '\n';
  for (auto& i : ans) cout << i << ' ';
  cout << '\n';
  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;
const int N = 1003;
int v[N][6];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 5; j++) cin >> v[i][j];
  if (n > 25) return puts("0"), 0;
  vector<int> ans;
  for (int i = 0; i < n; i++) {
    int cnt = 0;
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      for (int k = 0; k < n; k++) {
        if (k == i || k == j) continue;
        int cur = 0;
        for (int z = 0; z < 5; z++) {
          cur += (v[j][z] - v[i][z]) * (v[k][z] - v[i][z]);
        }
        cnt |= cur > 0;
      }
    }
    if (!cnt) ans.push_back(i + 1);
  }
  cout << ans.size() << '\n';
  for (auto& i : ans) cout << i << ' ';
  cout << '\n';
  return 0;
}

```