#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  long long scal;
  scanf("%d", &n);
  int a, b, c, d, e;
  vector<vector<int> > points;
  vector<int> good, v;
  for (int i = (0); i < (n); i++) {
    scanf("%d%d%d%d%d", &a, &b, &c, &d, &e);
    v = {a, b, c, d, e};
    points.push_back(v);
  }
  if (n > 12)
    cout << 0 << endl;
  else {
    for (int a = 0; a < n; a++) {
      bool ok = true;
      for (int b = 0; b < n && ok; b++) {
        for (int c = 0; c < n && ok; c++) {
          if (a != b && b != c && a != c) {
            scal =
                (points[b][4] - points[a][4]) * (points[c][4] - points[a][4]) +
                (points[b][3] - points[a][3]) * (points[c][3] - points[a][3]) +
                (points[b][2] - points[a][2]) * (points[c][2] - points[a][2]) +
                (points[b][1] - points[a][1]) * (points[c][1] - points[a][1]) +
                (points[b][0] - points[a][0]) * (points[c][0] - points[a][0]);
            if (scal > 0) ok = false;
          }
        }
      }
      if (ok) good.push_back(a + 1);
    }
    printf("%d\n", (int)good.size());
    for (int i = (0); i < (good.size()); i++) printf("%d\n", good[i]);
  }
  return 0;
}

### Prompt

Generate 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;
int main() {
  int n;
  long long scal;
  scanf("%d", &n);
  int a, b, c, d, e;
  vector<vector<int> > points;
  vector<int> good, v;
  for (int i = (0); i < (n); i++) {
    scanf("%d%d%d%d%d", &a, &b, &c, &d, &e);
    v = {a, b, c, d, e};
    points.push_back(v);
  }
  if (n > 12)
    cout << 0 << endl;
  else {
    for (int a = 0; a < n; a++) {
      bool ok = true;
      for (int b = 0; b < n && ok; b++) {
        for (int c = 0; c < n && ok; c++) {
          if (a != b && b != c && a != c) {
            scal =
                (points[b][4] - points[a][4]) * (points[c][4] - points[a][4]) +
                (points[b][3] - points[a][3]) * (points[c][3] - points[a][3]) +
                (points[b][2] - points[a][2]) * (points[c][2] - points[a][2]) +
                (points[b][1] - points[a][1]) * (points[c][1] - points[a][1]) +
                (points[b][0] - points[a][0]) * (points[c][0] - points[a][0]);
            if (scal > 0) ok = false;
          }
        }
      }
      if (ok) good.push_back(a + 1);
    }
    printf("%d\n", (int)good.size());
    for (int i = (0); i < (good.size()); i++) printf("%d\n", good[i]);
  }
  return 0;
}

```