#include <bits/stdc++.h>
using namespace std;
vector<vector<int> > pt;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  if (n > 300) {
    cout << 0 << endl;
    return 0;
  }
  pt.resize(n, vector<int>(5));
  for (int i = 0; i < (n); ++i) {
    for (int j = 0; j < (5); ++j) cin >> pt[i][j];
  }
  vector<int> gd;
  bool isg;
  int tpr;
  for (int i = 0; i < (n); ++i) {
    isg = 1;
    for (int j = 0; j < (n); ++j) {
      if (i == j) continue;
      for (int k = 0; k < (j); ++k) {
        if (i == k) continue;
        tpr = 0;
        for (int z = 0; z < (5); ++z) {
          tpr += (pt[j][z] - pt[i][z]) * (pt[k][z] - pt[i][z]);
        }
        if (tpr > 0) {
          isg = 0;
          break;
        }
      }
      if (!isg) break;
    }
    if (isg) gd.push_back(i + 1);
  }
  cout << (int((gd).size())) << endl;
  for (int i = 0; i < ((int((gd).size()))); ++i) cout << gd[i] << " ";
  cout << endl;
  return 0;
}

### Prompt

Create a solution in cpp for 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;
vector<vector<int> > pt;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  if (n > 300) {
    cout << 0 << endl;
    return 0;
  }
  pt.resize(n, vector<int>(5));
  for (int i = 0; i < (n); ++i) {
    for (int j = 0; j < (5); ++j) cin >> pt[i][j];
  }
  vector<int> gd;
  bool isg;
  int tpr;
  for (int i = 0; i < (n); ++i) {
    isg = 1;
    for (int j = 0; j < (n); ++j) {
      if (i == j) continue;
      for (int k = 0; k < (j); ++k) {
        if (i == k) continue;
        tpr = 0;
        for (int z = 0; z < (5); ++z) {
          tpr += (pt[j][z] - pt[i][z]) * (pt[k][z] - pt[i][z]);
        }
        if (tpr > 0) {
          isg = 0;
          break;
        }
      }
      if (!isg) break;
    }
    if (isg) gd.push_back(i + 1);
  }
  cout << (int((gd).size())) << endl;
  for (int i = 0; i < ((int((gd).size()))); ++i) cout << gd[i] << " ";
  cout << endl;
  return 0;
}

```