#include <bits/stdc++.h>
using namespace std;
const int M = 1000000007;
const int MM = 998244353;
const long double PI = acos(-1);
const long long INF = 2e18;
template <typename T, typename T1>
void amax(T &a, T1 b) {
  if (b > a) a = b;
}
template <typename T, typename T1>
void amin(T &a, T1 b) {
  if (b < a) a = b;
}
long long power(long long b, long long e) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b, e / 2);
  return power(b * b, e / 2);
}
long long power(long long b, long long e, long long m) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b % m, e / 2, m) % m;
  return power(b * b % m, e / 2, m);
}
long long modinv(long long a, long long m) { return power(a, m - 2, m); }
int TLE_TERROR() {
  long long n;
  cin >> n;
  long long a[n], b[n], c[n], d[n], e[n];
  long double val[n];
  long long va, vb, vc, vd, ve;
  vector<long long> ans;
  for (long long i = 0; i < n; i++) {
    cin >> a[i] >> b[i] >> c[i] >> d[i] >> e[i];
  }
  if (n > 100) {
    cout << 0;
    return 0;
  }
  for (long long i = 0; i < n; i++) {
    long long f = 0;
    for (long long j = 0; j < n; j++) {
      if (j != i) {
        for (long long k = 0; k < n; k++) {
          if (k != i && k != j) {
            va = (a[j] - a[i]) * (a[k] - a[i]);
            vb = (b[j] - b[i]) * (b[k] - b[i]);
            vc = (c[j] - c[i]) * (c[k] - c[i]);
            vd = (d[j] - d[i]) * (d[k] - d[i]);
            ve = (e[j] - e[i]) * (e[k] - e[i]);
            long long z = va + vb + vc + vd + ve;
            if (z > 0) {
              f = 1;
              break;
            }
          }
        }
      }
      if (f == 1) break;
    }
    if (f == 0) ans.push_back(i + 1);
  }
  cout << ans.size() << "\n";
  for (long long i = 0; i < ans.size(); i++) cout << ans[i] << "\n";
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int TESTS = 1;
  while (TESTS--) TLE_TERROR();
  return 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;
const int M = 1000000007;
const int MM = 998244353;
const long double PI = acos(-1);
const long long INF = 2e18;
template <typename T, typename T1>
void amax(T &a, T1 b) {
  if (b > a) a = b;
}
template <typename T, typename T1>
void amin(T &a, T1 b) {
  if (b < a) a = b;
}
long long power(long long b, long long e) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b, e / 2);
  return power(b * b, e / 2);
}
long long power(long long b, long long e, long long m) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b % m, e / 2, m) % m;
  return power(b * b % m, e / 2, m);
}
long long modinv(long long a, long long m) { return power(a, m - 2, m); }
int TLE_TERROR() {
  long long n;
  cin >> n;
  long long a[n], b[n], c[n], d[n], e[n];
  long double val[n];
  long long va, vb, vc, vd, ve;
  vector<long long> ans;
  for (long long i = 0; i < n; i++) {
    cin >> a[i] >> b[i] >> c[i] >> d[i] >> e[i];
  }
  if (n > 100) {
    cout << 0;
    return 0;
  }
  for (long long i = 0; i < n; i++) {
    long long f = 0;
    for (long long j = 0; j < n; j++) {
      if (j != i) {
        for (long long k = 0; k < n; k++) {
          if (k != i && k != j) {
            va = (a[j] - a[i]) * (a[k] - a[i]);
            vb = (b[j] - b[i]) * (b[k] - b[i]);
            vc = (c[j] - c[i]) * (c[k] - c[i]);
            vd = (d[j] - d[i]) * (d[k] - d[i]);
            ve = (e[j] - e[i]) * (e[k] - e[i]);
            long long z = va + vb + vc + vd + ve;
            if (z > 0) {
              f = 1;
              break;
            }
          }
        }
      }
      if (f == 1) break;
    }
    if (f == 0) ans.push_back(i + 1);
  }
  cout << ans.size() << "\n";
  for (long long i = 0; i < ans.size(); i++) cout << ans[i] << "\n";
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int TESTS = 1;
  while (TESTS--) TLE_TERROR();
  return 0;
}

```