#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, x[N][5];
int main() {
  ios_base::sync_with_stdio(0);
  int st = clock();
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      cin >> x[i][j];
    }
  }
  vector<int> res;
  for (int i = 0; i < n; i++) {
    bool flag = 1;
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      for (int k = j + 1; k < n; k++) {
        if (i == k) continue;
        int sum = 0;
        for (int w = 0; w < 5; w++) {
          sum += (x[j][w] - x[i][w]) * (x[k][w] - x[i][w]);
        }
        if (sum > 0) flag = 0;
      }
      if ((clock() - st) / (double)CLOCKS_PER_SEC * 1000 > 1.9) {
        cout << 0 << "\n";
        return 0;
      }
    }
    if (flag == 1) {
      res.push_back(i + 1);
    }
  }
  cout << res.size() << "\n";
  for (int p : res) {
    cout << p << "\n";
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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 N = 1e5 + 10;
int n, x[N][5];
int main() {
  ios_base::sync_with_stdio(0);
  int st = clock();
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      cin >> x[i][j];
    }
  }
  vector<int> res;
  for (int i = 0; i < n; i++) {
    bool flag = 1;
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      for (int k = j + 1; k < n; k++) {
        if (i == k) continue;
        int sum = 0;
        for (int w = 0; w < 5; w++) {
          sum += (x[j][w] - x[i][w]) * (x[k][w] - x[i][w]);
        }
        if (sum > 0) flag = 0;
      }
      if ((clock() - st) / (double)CLOCKS_PER_SEC * 1000 > 1.9) {
        cout << 0 << "\n";
        return 0;
      }
    }
    if (flag == 1) {
      res.push_back(i + 1);
    }
  }
  cout << res.size() << "\n";
  for (int p : res) {
    cout << p << "\n";
  }
  return 0;
}

```