#include <bits/stdc++.h>
using namespace std;
long long N;
long long P[2000][5];
long long A[5];
long long B[5];
long long PP;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N;
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < 5; j++) cin >> P[i][j];
  }
  if (N >= 20) {
    cout << 0 << endl;
    return 0;
  }
  if (N <= 2) {
    cout << N << "\n";
    for (int i = 1; i <= N; i++) cout << i << "\n";
    return 0;
  }
  for (int i = 0; i < N; i++) {
    bool flag = true;
    for (int j = 0; j < N; j++) {
      if (flag == true && j != i) {
        for (int k = 0; k < N; k++) {
          if (k != j && k != i) {
            PP = 0;
            for (int h = 0; h < 5; h++) {
              A[h] = P[j][h] - P[i][h];
              B[h] = P[k][h] - P[i][h];
              PP += A[h] * B[h];
            }
            if (PP > 0) {
              flag = false;
              break;
            }
          }
        }
      } else {
        if (flag == false) break;
      }
    }
    if (flag == true) {
      cout << 1 << "\n" << i + 1 << "\n";
      return 0;
    }
  }
  cout << 0 << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
long long N;
long long P[2000][5];
long long A[5];
long long B[5];
long long PP;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N;
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < 5; j++) cin >> P[i][j];
  }
  if (N >= 20) {
    cout << 0 << endl;
    return 0;
  }
  if (N <= 2) {
    cout << N << "\n";
    for (int i = 1; i <= N; i++) cout << i << "\n";
    return 0;
  }
  for (int i = 0; i < N; i++) {
    bool flag = true;
    for (int j = 0; j < N; j++) {
      if (flag == true && j != i) {
        for (int k = 0; k < N; k++) {
          if (k != j && k != i) {
            PP = 0;
            for (int h = 0; h < 5; h++) {
              A[h] = P[j][h] - P[i][h];
              B[h] = P[k][h] - P[i][h];
              PP += A[h] * B[h];
            }
            if (PP > 0) {
              flag = false;
              break;
            }
          }
        }
      } else {
        if (flag == false) break;
      }
    }
    if (flag == true) {
      cout << 1 << "\n" << i + 1 << "\n";
      return 0;
    }
  }
  cout << 0 << endl;
  return 0;
}

```