#include <bits/stdc++.h>
using namespace std;
vector<int> v;
int n;
int a[1111][5];
int coss(int i, int j, int k) {
  int ans = 0;
  for (int it = 0; it < 5; it++) {
    ans += (a[i][it] - a[j][it]) * (a[i][it] - a[k][it]);
  }
  return ans;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      cin >> a[i][j];
    }
  }
  int count = 0;
  for (int i = 0; i < n; i++) {
    bool ok = true;
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      for (int k = j + 1; k < n; k++) {
        if (k == i) continue;
        if (coss(i, j, k) > 0) ok = false;
      }
      if (!ok) break;
    }
    if (ok) {
      v.push_back(i + 1);
    }
  }
  cout << v.size() << endl;
  for (int i = 0; i < v.size(); i++) {
    cout << v[i] << endl;
  }
}

### 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;
vector<int> v;
int n;
int a[1111][5];
int coss(int i, int j, int k) {
  int ans = 0;
  for (int it = 0; it < 5; it++) {
    ans += (a[i][it] - a[j][it]) * (a[i][it] - a[k][it]);
  }
  return ans;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      cin >> a[i][j];
    }
  }
  int count = 0;
  for (int i = 0; i < n; i++) {
    bool ok = true;
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      for (int k = j + 1; k < n; k++) {
        if (k == i) continue;
        if (coss(i, j, k) > 0) ok = false;
      }
      if (!ok) break;
    }
    if (ok) {
      v.push_back(i + 1);
    }
  }
  cout << v.size() << endl;
  for (int i = 0; i < v.size(); i++) {
    cout << v[i] << endl;
  }
}

```