#include <bits/stdc++.h>
using namespace std;
bool is_accute(vector<int> a, vector<int> b, vector<int> c) {
  for (int i = 0; i < 5; ++i) b[i] -= a[i], c[i] -= a[i];
  int ans = 0;
  for (int i = 0; i < 5; ++i) ans += b[i] * c[i];
  return ans > 0;
}
bool is_good[2000] = {0};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  if (n > 100) cout << 0 << '\n', exit(0);
  vector<vector<int>> A(n, vector<int>(5));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 5; ++j) cin >> A[i][j];
  }
  int ans = n;
  for (int i = 0; i < n; ++i) {
    bool found = false;
    for (int j = 0; j < n and not found; ++j) {
      for (int k = 0; k < n and not found; ++k) {
        if (i == j or j == k or k == i) continue;
        found |= is_accute(A[i], A[j], A[k]);
      }
    }
    is_good[i] = found;
    ans -= found;
  }
  cout << ans << '\n';
  for (int i = 0; i < n; ++i)
    if (is_good[i] == 0) cout << i + 1 << '\n';
}

### Prompt

Please formulate 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;
bool is_accute(vector<int> a, vector<int> b, vector<int> c) {
  for (int i = 0; i < 5; ++i) b[i] -= a[i], c[i] -= a[i];
  int ans = 0;
  for (int i = 0; i < 5; ++i) ans += b[i] * c[i];
  return ans > 0;
}
bool is_good[2000] = {0};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  if (n > 100) cout << 0 << '\n', exit(0);
  vector<vector<int>> A(n, vector<int>(5));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 5; ++j) cin >> A[i][j];
  }
  int ans = n;
  for (int i = 0; i < n; ++i) {
    bool found = false;
    for (int j = 0; j < n and not found; ++j) {
      for (int k = 0; k < n and not found; ++k) {
        if (i == j or j == k or k == i) continue;
        found |= is_accute(A[i], A[j], A[k]);
      }
    }
    is_good[i] = found;
    ans -= found;
  }
  cout << ans << '\n';
  for (int i = 0; i < n; ++i)
    if (is_good[i] == 0) cout << i + 1 << '\n';
}

```