#include <bits/stdc++.h>
using namespace std;
int A[1005][5];
bool good[1005];
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, i, j, k, l, tmp, res;
  cin >> n;
  for (i = 1; i <= n; i++)
    for (j = 0; j < 5; j++) cin >> A[i][j];
  if (n > 11) {
    cout << 0 << '\n';
    return 0;
  }
  for (i = 1, res = 0; i <= n; i++) {
    good[i] = true;
    for (j = 1; j <= n; j++)
      for (k = 1; k <= n; k++)
        if (j != i && k != i && j != k) {
          for (l = tmp = 0; l < 5; l++)
            tmp += (A[j][l] - A[i][l]) * (A[k][l] - A[i][l]);
          if (tmp > 0) {
            good[i] = false;
            j = n + 1;
            break;
          }
        }
    if (good[i]) res++;
  }
  cout << res << '\n';
  for (i = 1; i <= n; i++)
    if (good[i]) cout << i << '\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 A[1005][5];
bool good[1005];
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, i, j, k, l, tmp, res;
  cin >> n;
  for (i = 1; i <= n; i++)
    for (j = 0; j < 5; j++) cin >> A[i][j];
  if (n > 11) {
    cout << 0 << '\n';
    return 0;
  }
  for (i = 1, res = 0; i <= n; i++) {
    good[i] = true;
    for (j = 1; j <= n; j++)
      for (k = 1; k <= n; k++)
        if (j != i && k != i && j != k) {
          for (l = tmp = 0; l < 5; l++)
            tmp += (A[j][l] - A[i][l]) * (A[k][l] - A[i][l]);
          if (tmp > 0) {
            good[i] = false;
            j = n + 1;
            break;
          }
        }
    if (good[i]) res++;
  }
  cout << res << '\n';
  for (i = 1; i <= n; i++)
    if (good[i]) cout << i << '\n';
  return 0;
}

```