#include <bits/stdc++.h>
using namespace std;
template <typename tn>
inline tn next(void) {
  tn k;
  cin >> k;
  return k;
}
template <typename tn>
inline ostream& operator<<(ostream& os, const vector<tn>& v) {
  for (int i = 0; (unsigned)i < v.size(); i++) os << v[i] << endl;
  return os;
}
struct point {
  int a[10];
  inline void input(void) {
    for (int i = 0; i < 5; i++) cin >> a[i];
  }
} w[1050];
vector<int> ans;
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T = next<int>();
  for (int i = 0; i < T; i++) w[i].input();
  for (int i = 0; i < T; i++) {
    bool Break = false;
    for (int j = 0; j < T && !Break; j++)
      for (int k = 0; k < T && !Break; k++) {
        int tmp = 0;
        if (i == j || j == k) continue;
        for (int l = 0; l < 5; l++)
          tmp += (w[j].a[l] - w[i].a[l]) * (w[k].a[l] - w[i].a[l]);
        if (tmp > 0) Break = 1;
      }
    if (!Break) ans.push_back(i + 1);
  }
  cout << ans.size() << endl << ans;
  return ~~(0 - 0);
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
template <typename tn>
inline tn next(void) {
  tn k;
  cin >> k;
  return k;
}
template <typename tn>
inline ostream& operator<<(ostream& os, const vector<tn>& v) {
  for (int i = 0; (unsigned)i < v.size(); i++) os << v[i] << endl;
  return os;
}
struct point {
  int a[10];
  inline void input(void) {
    for (int i = 0; i < 5; i++) cin >> a[i];
  }
} w[1050];
vector<int> ans;
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T = next<int>();
  for (int i = 0; i < T; i++) w[i].input();
  for (int i = 0; i < T; i++) {
    bool Break = false;
    for (int j = 0; j < T && !Break; j++)
      for (int k = 0; k < T && !Break; k++) {
        int tmp = 0;
        if (i == j || j == k) continue;
        for (int l = 0; l < 5; l++)
          tmp += (w[j].a[l] - w[i].a[l]) * (w[k].a[l] - w[i].a[l]);
        if (tmp > 0) Break = 1;
      }
    if (!Break) ans.push_back(i + 1);
  }
  cout << ans.size() << endl << ans;
  return ~~(0 - 0);
}

```