#include <bits/stdc++.h>
using namespace std;
int n, points[1009][5], ans = 0;
vector<int> good;
int dot(int a, int b, int c) {
  int res = 0;
  for (int d = 0; d < 5; d++) {
    res += (points[b][d] - points[a][d]) * (points[c][d] - points[a][d]);
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> points[i][0] >> points[i][1] >> points[i][2] >> points[i][3] >>
        points[i][4];
  }
  for (int i = 0; i < n; i++) {
    int check = 1;
    for (int j = 0; j < n; j++) {
      if (check) {
        for (int k = 0; k < n; k++) {
          if (i != j && j != k && i != k && dot(i, j, k) > 0) {
            check = 0;
            break;
          }
        }
      }
    }
    if (check) {
      ans++;
      good.push_back(i);
    }
  }
  cout << ans << endl;
  for (int x : good) {
    cout << x + 1 << " ";
  }
}

### Prompt

Your task is to create 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;
int n, points[1009][5], ans = 0;
vector<int> good;
int dot(int a, int b, int c) {
  int res = 0;
  for (int d = 0; d < 5; d++) {
    res += (points[b][d] - points[a][d]) * (points[c][d] - points[a][d]);
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> points[i][0] >> points[i][1] >> points[i][2] >> points[i][3] >>
        points[i][4];
  }
  for (int i = 0; i < n; i++) {
    int check = 1;
    for (int j = 0; j < n; j++) {
      if (check) {
        for (int k = 0; k < n; k++) {
          if (i != j && j != k && i != k && dot(i, j, k) > 0) {
            check = 0;
            break;
          }
        }
      }
    }
    if (check) {
      ans++;
      good.push_back(i);
    }
  }
  cout << ans << endl;
  for (int x : good) {
    cout << x + 1 << " ";
  }
}

```