#include <bits/stdc++.h>
using namespace std;
const int N = 2e3 + 123;
const int mod = 1e9 + 7;
const int INF = 1e9 + 1;
const long long INFL = 3e18 + 1;
const double pi = acos(-1.0);
inline void DoBetter() {
  unsigned int FOR;
  asm("rdtsc" : "=a"(FOR));
  srand(FOR);
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int n, a[N], b[N], c[N], d[N], e[N];
int pos[N], res;
int scalar(int i, int j, int k) {
  return (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]);
}
int main() {
  DoBetter();
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i] >> b[i] >> c[i] >> d[i] >> e[i];
  for (int i = 1; i <= n; i++) {
    bool ok = 0;
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      ok = 0;
      for (int k = 1; k <= n; k++) {
        if (k == j || k == i) continue;
        if (scalar(i, j, k) <= 0) continue;
        pos[i] = ok = 1;
        res++;
        break;
      }
      if (ok) break;
    }
  }
  cout << n - res << endl;
  for (int i = 1; i <= n; i++)
    if (pos[i] == 0) cout << i << ' ';
  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;
const int N = 2e3 + 123;
const int mod = 1e9 + 7;
const int INF = 1e9 + 1;
const long long INFL = 3e18 + 1;
const double pi = acos(-1.0);
inline void DoBetter() {
  unsigned int FOR;
  asm("rdtsc" : "=a"(FOR));
  srand(FOR);
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int n, a[N], b[N], c[N], d[N], e[N];
int pos[N], res;
int scalar(int i, int j, int k) {
  return (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]);
}
int main() {
  DoBetter();
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i] >> b[i] >> c[i] >> d[i] >> e[i];
  for (int i = 1; i <= n; i++) {
    bool ok = 0;
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      ok = 0;
      for (int k = 1; k <= n; k++) {
        if (k == j || k == i) continue;
        if (scalar(i, j, k) <= 0) continue;
        pos[i] = ok = 1;
        res++;
        break;
      }
      if (ok) break;
    }
  }
  cout << n - res << endl;
  for (int i = 1; i <= n; i++)
    if (pos[i] == 0) cout << i << ' ';
  return 0;
}

```