#include <bits/stdc++.h>
using namespace std;
void smain();
int main() {
  ios_base::sync_with_stdio(0);
  smain();
  return 0;
}
long long n;
long long a[1001][5];
vector<long long> solve() {
  long long m = min(n, 50LL);
  vector<long long> res;
  for (long long i = 0; i < n; i++) {
    bool ok = 1;
    for (long long j = 0; j < m; j++)
      if (ok && j != i)
        for (long long k = 0; k < j; k++)
          if (ok && k != i) {
            long long s = 0;
            for (long long l = 0; l < 5; l++)
              s += (a[j][l] - a[i][l]) * (a[k][l] - a[i][l]);
            if (s > 0) {
              ok = 0;
              break;
            }
          }
    if (ok) res.push_back(i);
  }
  return res;
}
void smain() {
  for (; cin >> n;) {
    for (long long i = 0; i < n; i++)
      for (long long j = 0; j < 5; j++) cin >> a[i][j];
    auto res = solve();
    cout << res.size();
    for (auto i : res) cout << '\n' << i + 1;
  }
}

### 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;
void smain();
int main() {
  ios_base::sync_with_stdio(0);
  smain();
  return 0;
}
long long n;
long long a[1001][5];
vector<long long> solve() {
  long long m = min(n, 50LL);
  vector<long long> res;
  for (long long i = 0; i < n; i++) {
    bool ok = 1;
    for (long long j = 0; j < m; j++)
      if (ok && j != i)
        for (long long k = 0; k < j; k++)
          if (ok && k != i) {
            long long s = 0;
            for (long long l = 0; l < 5; l++)
              s += (a[j][l] - a[i][l]) * (a[k][l] - a[i][l]);
            if (s > 0) {
              ok = 0;
              break;
            }
          }
    if (ok) res.push_back(i);
  }
  return res;
}
void smain() {
  for (; cin >> n;) {
    for (long long i = 0; i < n; i++)
      for (long long j = 0; j < 5; j++) cin >> a[i][j];
    auto res = solve();
    cout << res.size();
    for (auto i : res) cout << '\n' << i + 1;
  }
}

```