#include <bits/stdc++.h>
using namespace std;
const int N = 1005;
long long a[N], b[N], c[N], d[N], e[N];
long long t1[N], t2[N], t3[N], t4[N], t5[N];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i] >> b[i] >> c[i] >> d[i] >> e[i];
  int sum = 0;
  int ans[N];
  if (n >= 30)
    cout << "0" << endl;
  else {
    for (int i = 0; i < n; i++) {
      bool flag = true;
      for (int j = 0; j < n; j++) {
        for (int k = 0; k < n; k++) {
          if (j != i && k != i && k != j) {
            if ((a[j] - a[i]) * (a[k] - a[i]) + (b[j] - b[i]) * (b[k] - b[i]) +
                    (c[j] - c[i]) * (c[k] - c[i]) +
                    (d[j] - d[i]) * (d[k] - d[i]) +
                    (e[j] - e[i]) * (e[k] - e[i]) >
                0)
              flag = false;
          }
        }
      }
      if (flag) ans[sum++] = i + 1;
    }
    cout << sum << endl;
    if (sum > 0) {
      for (int i = 0; i < sum; i++) cout << ans[i] << ' ';
      cout << endl;
    }
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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 = 1005;
long long a[N], b[N], c[N], d[N], e[N];
long long t1[N], t2[N], t3[N], t4[N], t5[N];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i] >> b[i] >> c[i] >> d[i] >> e[i];
  int sum = 0;
  int ans[N];
  if (n >= 30)
    cout << "0" << endl;
  else {
    for (int i = 0; i < n; i++) {
      bool flag = true;
      for (int j = 0; j < n; j++) {
        for (int k = 0; k < n; k++) {
          if (j != i && k != i && k != j) {
            if ((a[j] - a[i]) * (a[k] - a[i]) + (b[j] - b[i]) * (b[k] - b[i]) +
                    (c[j] - c[i]) * (c[k] - c[i]) +
                    (d[j] - d[i]) * (d[k] - d[i]) +
                    (e[j] - e[i]) * (e[k] - e[i]) >
                0)
              flag = false;
          }
        }
      }
      if (flag) ans[sum++] = i + 1;
    }
    cout << sum << endl;
    if (sum > 0) {
      for (int i = 0; i < sum; i++) cout << ans[i] << ' ';
      cout << endl;
    }
  }
  return 0;
}

```