#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 9;
const long long int MOD = 1e9 + 696969;
const long long INF = 1e18 + 3;
int tab[1010][5];
long long int x[5], y[5];
bool spoko[1010];
int n;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = (1); i <= (n); ++i) {
    for (int dim = (0); dim <= (4); ++dim) cin >> tab[i][dim];
  }
  for (int i = (1); i <= (n); ++i) spoko[i] = 1;
  for (int i = 1; i <= n; ++i) {
    if (!spoko[i]) continue;
    for (int j = 1; j <= n; ++j) {
      if (!spoko[i]) break;
      for (int k = 1; k <= n; ++k) {
        if (i == j || i == k || k == j) continue;
        for (int h = (0); h <= (4); ++h) x[h] = tab[i][h] - tab[j][h];
        long long int res = 0;
        for (int h = (0); h <= (4); ++h) {
          y[h] = tab[k][h] - tab[i][h];
          res += x[h] * y[h];
        }
        if (res < 0) {
          spoko[i] = 0;
          break;
        }
      }
    }
  }
  int wskaz = 0;
  for (int i = (1); i <= (n); ++i)
    if (spoko[i]) ++wskaz;
  cout << wskaz << endl;
  for (int i = (1); i <= (n); ++i)
    if (spoko[i]) cout << i << ' ';
}

### Prompt

Your task is to create 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 int inf = 1e9 + 9;
const long long int MOD = 1e9 + 696969;
const long long INF = 1e18 + 3;
int tab[1010][5];
long long int x[5], y[5];
bool spoko[1010];
int n;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = (1); i <= (n); ++i) {
    for (int dim = (0); dim <= (4); ++dim) cin >> tab[i][dim];
  }
  for (int i = (1); i <= (n); ++i) spoko[i] = 1;
  for (int i = 1; i <= n; ++i) {
    if (!spoko[i]) continue;
    for (int j = 1; j <= n; ++j) {
      if (!spoko[i]) break;
      for (int k = 1; k <= n; ++k) {
        if (i == j || i == k || k == j) continue;
        for (int h = (0); h <= (4); ++h) x[h] = tab[i][h] - tab[j][h];
        long long int res = 0;
        for (int h = (0); h <= (4); ++h) {
          y[h] = tab[k][h] - tab[i][h];
          res += x[h] * y[h];
        }
        if (res < 0) {
          spoko[i] = 0;
          break;
        }
      }
    }
  }
  int wskaz = 0;
  for (int i = (1); i <= (n); ++i)
    if (spoko[i]) ++wskaz;
  cout << wskaz << endl;
  for (int i = (1); i <= (n); ++i)
    if (spoko[i]) cout << i << ' ';
}

```