#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
int t;
void solve();
struct Node {
  long long q, w, e, r, t;
};
int n;
vector<Node> a;
vector<bool> ans;
int res = 0;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  t = 1;
  for (int i = 0; i < t; ++i) {
    solve();
  }
}
void solve() {
  cin >> n;
  a.resize(n);
  ans.resize(n);
  for (int i = 0; i < int(n); ++i) {
    cin >> a[i].q >> a[i].w >> a[i].e >> a[i].r >> a[i].t;
  }
  if (n > 66) {
    cout << 0;
    return;
  }
  for (int i = 0; i < int(n); ++i) {
    for (int j = (0); j <= int(n - 1); ++j) {
      for (int k = (j + 1); k <= int(n - 1); ++k) {
        if (j == i || k == i) {
          continue;
        }
        if ((a[j].q - a[i].q) * (a[k].q - a[i].q) +
                (a[j].w - a[i].w) * (a[k].w - a[i].w) +
                (a[j].e - a[i].e) * (a[k].e - a[i].e) +
                (a[j].r - a[i].r) * (a[k].r - a[i].r) +
                (a[j].t - a[i].t) * (a[k].t - a[i].t) >
            0) {
          ans[i] = true;
          break;
        }
      }
      if (ans[i]) break;
    }
    if (!ans[i]) res++;
  }
  cout << res << endl;
  for (int i = 0; i < int(n); ++i) {
    if (!ans[i]) {
      cout << i + 1 << endl;
    }
  }
}

### Prompt

Your challenge is to write 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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
int t;
void solve();
struct Node {
  long long q, w, e, r, t;
};
int n;
vector<Node> a;
vector<bool> ans;
int res = 0;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  t = 1;
  for (int i = 0; i < t; ++i) {
    solve();
  }
}
void solve() {
  cin >> n;
  a.resize(n);
  ans.resize(n);
  for (int i = 0; i < int(n); ++i) {
    cin >> a[i].q >> a[i].w >> a[i].e >> a[i].r >> a[i].t;
  }
  if (n > 66) {
    cout << 0;
    return;
  }
  for (int i = 0; i < int(n); ++i) {
    for (int j = (0); j <= int(n - 1); ++j) {
      for (int k = (j + 1); k <= int(n - 1); ++k) {
        if (j == i || k == i) {
          continue;
        }
        if ((a[j].q - a[i].q) * (a[k].q - a[i].q) +
                (a[j].w - a[i].w) * (a[k].w - a[i].w) +
                (a[j].e - a[i].e) * (a[k].e - a[i].e) +
                (a[j].r - a[i].r) * (a[k].r - a[i].r) +
                (a[j].t - a[i].t) * (a[k].t - a[i].t) >
            0) {
          ans[i] = true;
          break;
        }
      }
      if (ans[i]) break;
    }
    if (!ans[i]) res++;
  }
  cout << res << endl;
  for (int i = 0; i < int(n); ++i) {
    if (!ans[i]) {
      cout << i + 1 << endl;
    }
  }
}

```