#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  using Point = array<int, 5>;
  vector<Point> points(n);
  for (Point& p : points)
    for (int i = 0; i < 5; i++) cin >> p[i];
  vector<int> good_indices;
  for (int i = 0; i < n; i++) {
    Point p = points[i];
    vector<Point> translates;
    for (int j = 0; j < n; j++)
      if (j != i) {
        Point q = points[j];
        for (int k = 0; k < 5; k++) q[k] -= p[k];
        translates.push_back(q);
      }
    bool bad = false;
    for (int j = 0; j < n - 1; j++) {
      if (bad) break;
      for (int k = j + 1; k < n - 1; k++)
        if (translates[j][0] * translates[k][0] +
                translates[j][1] * translates[k][1] +
                translates[j][2] * translates[k][2] +
                translates[j][3] * translates[k][3] +
                translates[j][4] * translates[k][4] >
            0) {
          bad = true;
          break;
        }
    }
    if (!bad) good_indices.push_back(i);
  }
  cout << good_indices.size() << endl;
  for (const int i : good_indices) cout << i + 1 << "\n";
}

### 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() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  using Point = array<int, 5>;
  vector<Point> points(n);
  for (Point& p : points)
    for (int i = 0; i < 5; i++) cin >> p[i];
  vector<int> good_indices;
  for (int i = 0; i < n; i++) {
    Point p = points[i];
    vector<Point> translates;
    for (int j = 0; j < n; j++)
      if (j != i) {
        Point q = points[j];
        for (int k = 0; k < 5; k++) q[k] -= p[k];
        translates.push_back(q);
      }
    bool bad = false;
    for (int j = 0; j < n - 1; j++) {
      if (bad) break;
      for (int k = j + 1; k < n - 1; k++)
        if (translates[j][0] * translates[k][0] +
                translates[j][1] * translates[k][1] +
                translates[j][2] * translates[k][2] +
                translates[j][3] * translates[k][3] +
                translates[j][4] * translates[k][4] >
            0) {
          bad = true;
          break;
        }
    }
    if (!bad) good_indices.push_back(i);
  }
  cout << good_indices.size() << endl;
  for (const int i : good_indices) cout << i + 1 << "\n";
}

```