#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  vector<vector<int>> X(n, vector<int>(5));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      cin >> X[i][j];
    }
  }
  if (n > 100) {
    cout << 0 << endl;
  } else {
    vector<bool> good(n, true);
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        for (int k = 0; k < n; k++) {
          if (i != j && j != k && k != i) {
            int dot = 0;
            for (int l = 0; l < 5; l++) {
              dot += (X[j][l] - X[i][l]) * (X[k][l] - X[i][l]);
            }
            if (dot > 0) {
              good[i] = false;
            }
          }
        }
      }
    }
    vector<int> ans;
    for (int i = 0; i < n; i++) {
      if (good[i]) {
        ans.push_back(i);
      }
    }
    int k = ans.size();
    cout << k << endl;
    for (int i = 0; i < k; i++) {
      cout << ans[i] + 1 << endl;
    }
  }
}

### Prompt

In cpp, your task is to solve 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() {
  int n;
  cin >> n;
  vector<vector<int>> X(n, vector<int>(5));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      cin >> X[i][j];
    }
  }
  if (n > 100) {
    cout << 0 << endl;
  } else {
    vector<bool> good(n, true);
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        for (int k = 0; k < n; k++) {
          if (i != j && j != k && k != i) {
            int dot = 0;
            for (int l = 0; l < 5; l++) {
              dot += (X[j][l] - X[i][l]) * (X[k][l] - X[i][l]);
            }
            if (dot > 0) {
              good[i] = false;
            }
          }
        }
      }
    }
    vector<int> ans;
    for (int i = 0; i < n; i++) {
      if (good[i]) {
        ans.push_back(i);
      }
    }
    int k = ans.size();
    cout << k << endl;
    for (int i = 0; i < k; i++) {
      cout << ans[i] + 1 << endl;
    }
  }
}

```