#include <bits/stdc++.h>
using namespace std;
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  int n;
  cin >> n;
  if (n > 100) {
    cout << 0;
    return 0;
  }
  vector<vector<int>> pts(n, vector<int>(5));
  for (vector<int> &v : pts) {
    for (int &i : v) cin >> i;
  }
  vector<int> ans;
  for (int a = 0; a < n; a++) {
    bool good = true;
    for (int b = 0; b < n && good; b++) {
      if (a == b) continue;
      for (int c = 0; c < n; c++) {
        if (a == c || b == c) continue;
        int dot = 0;
        for (int i = 0; i < 5; i++) {
          dot += (pts[b][i] - pts[a][i]) * (pts[c][i] - pts[a][i]);
        }
        if (dot > 0) {
          good = false;
          break;
        }
      }
    }
    if (good) {
      ans.push_back(a + 1);
    }
  }
  cout << ans.size() << "\n";
  for (int i : ans) cout << i << "\n";
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  int n;
  cin >> n;
  if (n > 100) {
    cout << 0;
    return 0;
  }
  vector<vector<int>> pts(n, vector<int>(5));
  for (vector<int> &v : pts) {
    for (int &i : v) cin >> i;
  }
  vector<int> ans;
  for (int a = 0; a < n; a++) {
    bool good = true;
    for (int b = 0; b < n && good; b++) {
      if (a == b) continue;
      for (int c = 0; c < n; c++) {
        if (a == c || b == c) continue;
        int dot = 0;
        for (int i = 0; i < 5; i++) {
          dot += (pts[b][i] - pts[a][i]) * (pts[c][i] - pts[a][i]);
        }
        if (dot > 0) {
          good = false;
          break;
        }
      }
    }
    if (good) {
      ans.push_back(a + 1);
    }
  }
  cout << ans.size() << "\n";
  for (int i : ans) cout << i << "\n";
}

```