#include <bits/stdc++.h>
using namespace std;
const int N = 1024;
int crd[N][5], good[N];
int main() {
  int n;
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 5; ++j) {
      cin >> crd[i][j];
    }
  }
  if (n > 100) {
    cout << 0 << endl;
  } else {
    int rcnt = 0;
    for (int i = 0; i < n; ++i) {
      good[i] = 1;
      for (int j = 0; j < n && good[i]; ++j)
        for (int k = j + 1; k < n; ++k) {
          if (j != i && k != i) {
            int dp = 0;
            for (int l = 0; l < 5; ++l)
              dp += (crd[j][l] - crd[i][l]) * (crd[k][l] - crd[i][l]);
            if (dp > 0) {
              good[i] = 0;
              break;
            }
          }
        }
      rcnt += good[i];
    }
    cout << rcnt << endl;
    for (int l = 0; rcnt && l < n; ++l) {
      if (good[l]) {
        cout << l + 1;
        if (--rcnt)
          cout << ' ';
        else {
          cout << endl;
        }
      }
    }
  }
  return 0;
}

### Prompt

Please formulate 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;
const int N = 1024;
int crd[N][5], good[N];
int main() {
  int n;
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 5; ++j) {
      cin >> crd[i][j];
    }
  }
  if (n > 100) {
    cout << 0 << endl;
  } else {
    int rcnt = 0;
    for (int i = 0; i < n; ++i) {
      good[i] = 1;
      for (int j = 0; j < n && good[i]; ++j)
        for (int k = j + 1; k < n; ++k) {
          if (j != i && k != i) {
            int dp = 0;
            for (int l = 0; l < 5; ++l)
              dp += (crd[j][l] - crd[i][l]) * (crd[k][l] - crd[i][l]);
            if (dp > 0) {
              good[i] = 0;
              break;
            }
          }
        }
      rcnt += good[i];
    }
    cout << rcnt << endl;
    for (int l = 0; rcnt && l < n; ++l) {
      if (good[l]) {
        cout << l + 1;
        if (--rcnt)
          cout << ' ';
        else {
          cout << endl;
        }
      }
    }
  }
  return 0;
}

```