#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  long long a[n + 1][6];
  for (int i = 0; i < n; i++) {
    for (int j = 1; j <= 5; j++) {
      cin >> a[i][j];
    }
  }
  vector<int> ans;
  for (int i = 0; i < n; i++) {
    int cnt = 0;
    bool flag = true;
    for (int l = 1; l < n; l++) {
      if (!flag) {
        break;
      }
      long long v[6], vv[6], den1 = 0;
      long double d1 = 0.0;
      for (int j = 1; j <= 5; j++) {
        v[j] = a[(i + l) % n][j] - a[i][j];
        den1 = den1 + (v[j] * v[j]);
      }
      d1 = sqrt(den1);
      for (int k = 1; k < n; k++) {
        if (k == l) {
          continue;
        }
        int curr = i + k;
        curr %= n;
        for (int j = 1; j <= 5; j++) {
          vv[j] = a[curr][j] - a[i][j];
        }
        long long num = 0, den2 = 0;
        long double d2 = 0.0;
        for (int j = 1; j <= 5; j++) {
          num = num + (v[j] * vv[j]);
          den2 = den2 + (vv[j] * vv[j]);
        }
        if (num > 0) {
          flag = false;
          break;
        }
      }
      cnt++;
    }
    if (flag) {
      ans.push_back(i);
    }
  }
  cout << ans.size() << "\n";
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i] + 1 << "\n";
  }
}

### Prompt

Generate 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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  long long a[n + 1][6];
  for (int i = 0; i < n; i++) {
    for (int j = 1; j <= 5; j++) {
      cin >> a[i][j];
    }
  }
  vector<int> ans;
  for (int i = 0; i < n; i++) {
    int cnt = 0;
    bool flag = true;
    for (int l = 1; l < n; l++) {
      if (!flag) {
        break;
      }
      long long v[6], vv[6], den1 = 0;
      long double d1 = 0.0;
      for (int j = 1; j <= 5; j++) {
        v[j] = a[(i + l) % n][j] - a[i][j];
        den1 = den1 + (v[j] * v[j]);
      }
      d1 = sqrt(den1);
      for (int k = 1; k < n; k++) {
        if (k == l) {
          continue;
        }
        int curr = i + k;
        curr %= n;
        for (int j = 1; j <= 5; j++) {
          vv[j] = a[curr][j] - a[i][j];
        }
        long long num = 0, den2 = 0;
        long double d2 = 0.0;
        for (int j = 1; j <= 5; j++) {
          num = num + (v[j] * vv[j]);
          den2 = den2 + (vv[j] * vv[j]);
        }
        if (num > 0) {
          flag = false;
          break;
        }
      }
      cnt++;
    }
    if (flag) {
      ans.push_back(i);
    }
  }
  cout << ans.size() << "\n";
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i] + 1 << "\n";
  }
}

```