#include <bits/stdc++.h>
using namespace std;
int in[1050][5];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, i, j, k, l;
  cin >> n;
  for (int i = (1); i <= (n); ++i)
    for (int j = 0; j < (5); ++j) cin >> in[i][j];
  if (n >= 20) {
    cout << "0";
    return 0;
  }
  vector<int> v;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      for (k = j + 1; k <= n; k++) {
        if (i == j || i == k) continue;
        long long s = 0;
        for (int l = 0; l < (5); ++l)
          s += (in[j][l] - in[i][l]) * (in[k][l] - in[i][l]);
        if (s > 0) break;
      }
      if (k <= n) break;
    }
    if (j > n) v.push_back(i);
  }
  cout << v.size() << endl;
  for (int i = 0; i < (v.size()); ++i) cout << v[i] << endl;
  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 in[1050][5];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, i, j, k, l;
  cin >> n;
  for (int i = (1); i <= (n); ++i)
    for (int j = 0; j < (5); ++j) cin >> in[i][j];
  if (n >= 20) {
    cout << "0";
    return 0;
  }
  vector<int> v;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      for (k = j + 1; k <= n; k++) {
        if (i == j || i == k) continue;
        long long s = 0;
        for (int l = 0; l < (5); ++l)
          s += (in[j][l] - in[i][l]) * (in[k][l] - in[i][l]);
        if (s > 0) break;
      }
      if (k <= n) break;
    }
    if (j > n) v.push_back(i);
  }
  cout << v.size() << endl;
  for (int i = 0; i < (v.size()); ++i) cout << v[i] << endl;
  return 0;
}

```