#include <bits/stdc++.h>
using namespace std;
int a[11][5];
bool ok(int l, int j, int k) {
  int i, an = 0;
  for (i = 0; i < 5; i++) {
    an = an + (a[j][i] - a[l][i]) * (a[k][i] - a[l][i]);
  }
  if (an > 0) return false;
  return true;
}
int main() {
  ios::sync_with_stdio(false);
  int i, j, k, n;
  cin >> n;
  if (n > 11) {
    n *= 5;
    while (n--) cin >> i;
    cout << 0;
  } else {
    for (i = 0; i < n; i++)
      cin >> a[i][0] >> a[i][1] >> a[i][2] >> a[i][3] >> a[i][4];
    vector<int> v;
    k = n;
    j = n;
    for (i = 0; i < n; i++) {
      k = n;
      for (j = 0; j < n && k == n; j++)
        for (k = j + 1; k < n; k++) {
          if (!ok(i, j, k)) {
            break;
          }
        }
      if (j == n && k == n) v.push_back(i + 1);
    }
    cout << v.size() << endl;
    for (i = 0; i < v.size(); i++) cout << v[i] << ' ' << endl;
    ;
  }
  return 0;
}

### Prompt

Your task is to create 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 a[11][5];
bool ok(int l, int j, int k) {
  int i, an = 0;
  for (i = 0; i < 5; i++) {
    an = an + (a[j][i] - a[l][i]) * (a[k][i] - a[l][i]);
  }
  if (an > 0) return false;
  return true;
}
int main() {
  ios::sync_with_stdio(false);
  int i, j, k, n;
  cin >> n;
  if (n > 11) {
    n *= 5;
    while (n--) cin >> i;
    cout << 0;
  } else {
    for (i = 0; i < n; i++)
      cin >> a[i][0] >> a[i][1] >> a[i][2] >> a[i][3] >> a[i][4];
    vector<int> v;
    k = n;
    j = n;
    for (i = 0; i < n; i++) {
      k = n;
      for (j = 0; j < n && k == n; j++)
        for (k = j + 1; k < n; k++) {
          if (!ok(i, j, k)) {
            break;
          }
        }
      if (j == n && k == n) v.push_back(i + 1);
    }
    cout << v.size() << endl;
    for (i = 0; i < v.size(); i++) cout << v[i] << ' ' << endl;
    ;
  }
  return 0;
}

```