#include <bits/stdc++.h>
using namespace std;
int n;
long long sp, t[1010][6], v1[6], v2[6];
vector<int> ans;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  if (n > 100) {
    cout << 0 << '\n';
    return 0;
  }
  for (int a = 1; a <= n; a++) {
    for (int b = 0; b < 5; b++) cin >> t[a][b];
  }
  for (int b = 1; b <= n; b++) {
    bool blad = 0;
    for (int c = 1; c <= n; c++) {
      for (int d = 1; d <= n; d++) {
        if (b != c && c != d && b != d) {
          for (int e = 0; e < 5; e++) v1[e] = t[d][e] - t[b][e];
          for (int e = 0; e < 5; e++) v2[e] = t[c][e] - t[b][e];
          sp = 0;
          for (int e = 0; e < 5; e++) sp += v1[e] * v2[e];
          if (sp > 0) blad = 1;
        }
      }
    }
    if (!blad) ans.push_back(b);
  }
  cout << ans.size() << '\n';
  for (int a = 0; a < ans.size(); a++) cout << ans[a] << '\n';
  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;
int n;
long long sp, t[1010][6], v1[6], v2[6];
vector<int> ans;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  if (n > 100) {
    cout << 0 << '\n';
    return 0;
  }
  for (int a = 1; a <= n; a++) {
    for (int b = 0; b < 5; b++) cin >> t[a][b];
  }
  for (int b = 1; b <= n; b++) {
    bool blad = 0;
    for (int c = 1; c <= n; c++) {
      for (int d = 1; d <= n; d++) {
        if (b != c && c != d && b != d) {
          for (int e = 0; e < 5; e++) v1[e] = t[d][e] - t[b][e];
          for (int e = 0; e < 5; e++) v2[e] = t[c][e] - t[b][e];
          sp = 0;
          for (int e = 0; e < 5; e++) sp += v1[e] * v2[e];
          if (sp > 0) blad = 1;
        }
      }
    }
    if (!blad) ans.push_back(b);
  }
  cout << ans.size() << '\n';
  for (int a = 0; a < ans.size(); a++) cout << ans[a] << '\n';
  return 0;
}

```