#include <bits/stdc++.h>
using namespace std;
struct point {
  int a, b, c, d, e;
};
int N;
vector<int> v;
const long double PI = acos(0);
int dot(point a, point b) {
  return a.a * b.a + a.b * b.b + a.c * b.c + a.d * b.d + a.e * b.e;
}
point sub(point a, point b) {
  return {b.a - a.a, b.b - a.b, b.c - a.c, b.d - a.d, b.e - a.e};
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N;
  if (N > 50) {
    cout << 0 << "\n";
    return 0;
  }
  vector<point> v;
  v.push_back({0, 0, 0, 0, 0});
  for (int i = 1; i <= N; i++) {
    point p;
    cin >> p.a >> p.b >> p.c >> p.d >> p.e;
    v.push_back(p);
  }
  vector<int> ans;
  for (int i = 1; i <= N; i++) {
    bool b = 1;
    for (int j = 1; j <= N; j++) {
      for (int k = 1; k <= N; k++) {
        if (j != k && i != j && i != k) {
          auto p = sub(v[i], v[j]), q = sub(v[i], v[k]);
          if (acos(dot(p, q) / sqrt(dot(p, p)) / sqrt(dot(q, q))) < PI) {
            b = 0;
          }
        }
      }
    }
    if (b) {
      ans.push_back(i);
    }
  }
  cout << ans.size() << "\n";
  for (int n : ans) {
    cout << n << "\n";
  }
}

### 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;
struct point {
  int a, b, c, d, e;
};
int N;
vector<int> v;
const long double PI = acos(0);
int dot(point a, point b) {
  return a.a * b.a + a.b * b.b + a.c * b.c + a.d * b.d + a.e * b.e;
}
point sub(point a, point b) {
  return {b.a - a.a, b.b - a.b, b.c - a.c, b.d - a.d, b.e - a.e};
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N;
  if (N > 50) {
    cout << 0 << "\n";
    return 0;
  }
  vector<point> v;
  v.push_back({0, 0, 0, 0, 0});
  for (int i = 1; i <= N; i++) {
    point p;
    cin >> p.a >> p.b >> p.c >> p.d >> p.e;
    v.push_back(p);
  }
  vector<int> ans;
  for (int i = 1; i <= N; i++) {
    bool b = 1;
    for (int j = 1; j <= N; j++) {
      for (int k = 1; k <= N; k++) {
        if (j != k && i != j && i != k) {
          auto p = sub(v[i], v[j]), q = sub(v[i], v[k]);
          if (acos(dot(p, q) / sqrt(dot(p, p)) / sqrt(dot(q, q))) < PI) {
            b = 0;
          }
        }
      }
    }
    if (b) {
      ans.push_back(i);
    }
  }
  cout << ans.size() << "\n";
  for (int n : ans) {
    cout << n << "\n";
  }
}

```