#include <bits/stdc++.h>
using namespace std;
const string __sep = "\n";
const int INF = 1000000001;
const int MOD = 998244353;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  t = 1;
  while (t-- > 0) {
    int n;
    cin >> n;
    vector<vector<int>> x(n);
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < 5; ++j) {
        int a;
        cin >> a;
        x[i].push_back(a);
      }
    }
    vector<int> good_;
    for (int i = 0; i < n; ++i) {
      bool good = true;
      for (int j = 0; j < n; ++j) {
        if (j == i) continue;
        for (int l = j + 1; l < n; ++l) {
          if (l == i) continue;
          int sum = 0;
          for (int k = 0; k < 5; ++k) {
            sum += (x[j][k] - x[i][k]) * (x[l][k] - x[i][k]);
          }
          if (sum > 0) {
            good = false;
            break;
          }
        }
        if (!good) break;
      }
      if (good) {
        good_.push_back(i);
      }
    }
    cout << good_.size() << endl;
    for (int s : good_) {
      cout << (1 + s) << "\n";
    }
  }
  return 0;
}

### 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;
const string __sep = "\n";
const int INF = 1000000001;
const int MOD = 998244353;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  t = 1;
  while (t-- > 0) {
    int n;
    cin >> n;
    vector<vector<int>> x(n);
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < 5; ++j) {
        int a;
        cin >> a;
        x[i].push_back(a);
      }
    }
    vector<int> good_;
    for (int i = 0; i < n; ++i) {
      bool good = true;
      for (int j = 0; j < n; ++j) {
        if (j == i) continue;
        for (int l = j + 1; l < n; ++l) {
          if (l == i) continue;
          int sum = 0;
          for (int k = 0; k < 5; ++k) {
            sum += (x[j][k] - x[i][k]) * (x[l][k] - x[i][k]);
          }
          if (sum > 0) {
            good = false;
            break;
          }
        }
        if (!good) break;
      }
      if (good) {
        good_.push_back(i);
      }
    }
    cout << good_.size() << endl;
    for (int s : good_) {
      cout << (1 + s) << "\n";
    }
  }
  return 0;
}

```