#include <bits/stdc++.h>
int n, a[1007], b[1007], c[1007], d[1007], e[1007];
std::vector<int> ans;
int main() {
  std::cin >> n;
  for (int i = 0; i < n; ++i) {
    std::cin >> a[i] >> b[i] >> c[i] >> d[i] >> e[i];
  }
  for (int i = 0; i < n; ++i) {
    bool ok = true;
    for (int j = 0; j < n; ++j) {
      if (j == i) continue;
      for (int k = 0; k < n; ++k) {
        if (k == i || k == j) continue;
        double x =
            (a[j] - a[i]) * (a[k] - a[i]) + (b[j] - b[i]) * (b[k] - b[i]) +
            (c[j] - c[i]) * (c[k] - c[i]) + (d[j] - d[i]) * (d[k] - d[i]) +
            (e[j] - e[i]) * (e[k] - e[i]);
        double y =
            sqrt((double)(a[j] - a[i]) * (a[j] - a[i]) +
                 (b[j] - b[i]) * (b[j] - b[i]) + (c[j] - c[i]) * (c[j] - c[i]) +
                 (d[j] - d[i]) * (d[j] - d[i]) + (e[j] - e[i]) * (e[j] - e[i]));
        double z =
            sqrt((double)(a[k] - a[i]) * (a[k] - a[i]) +
                 (b[k] - b[i]) * (b[k] - b[i]) + (c[k] - c[i]) * (c[k] - c[i]) +
                 (d[k] - d[i]) * (d[k] - d[i]) + (e[k] - e[i]) * (e[k] - e[i]));
        double alf = acos(x / y / z) * 180.0 / 3.14159265;
        if (alf < 90) {
          ok = false;
          break;
        }
      }
      if (!ok) break;
    }
    if (ok) {
      ans.push_back(i + 1);
    }
  }
  std::cout << ans.size() << "\n";
  for (auto el : ans) {
    std::cout << el << " ";
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
int n, a[1007], b[1007], c[1007], d[1007], e[1007];
std::vector<int> ans;
int main() {
  std::cin >> n;
  for (int i = 0; i < n; ++i) {
    std::cin >> a[i] >> b[i] >> c[i] >> d[i] >> e[i];
  }
  for (int i = 0; i < n; ++i) {
    bool ok = true;
    for (int j = 0; j < n; ++j) {
      if (j == i) continue;
      for (int k = 0; k < n; ++k) {
        if (k == i || k == j) continue;
        double x =
            (a[j] - a[i]) * (a[k] - a[i]) + (b[j] - b[i]) * (b[k] - b[i]) +
            (c[j] - c[i]) * (c[k] - c[i]) + (d[j] - d[i]) * (d[k] - d[i]) +
            (e[j] - e[i]) * (e[k] - e[i]);
        double y =
            sqrt((double)(a[j] - a[i]) * (a[j] - a[i]) +
                 (b[j] - b[i]) * (b[j] - b[i]) + (c[j] - c[i]) * (c[j] - c[i]) +
                 (d[j] - d[i]) * (d[j] - d[i]) + (e[j] - e[i]) * (e[j] - e[i]));
        double z =
            sqrt((double)(a[k] - a[i]) * (a[k] - a[i]) +
                 (b[k] - b[i]) * (b[k] - b[i]) + (c[k] - c[i]) * (c[k] - c[i]) +
                 (d[k] - d[i]) * (d[k] - d[i]) + (e[k] - e[i]) * (e[k] - e[i]));
        double alf = acos(x / y / z) * 180.0 / 3.14159265;
        if (alf < 90) {
          ok = false;
          break;
        }
      }
      if (!ok) break;
    }
    if (ok) {
      ans.push_back(i + 1);
    }
  }
  std::cout << ans.size() << "\n";
  for (auto el : ans) {
    std::cout << el << " ";
  }
}

```