#include <bits/stdc++.h>
using namespace std;
const int N = 1007;
const long double pi = acos(-1);
const long double eps = 1e-9;
int n;
int a[N];
int b[N];
int c[N];
int d[N];
int e[N];
vector<int> p;
bitset<N> zly;
long double len(long double a, long double b, long double c, long double d,
                long double e) {
  return sqrt(a * a + b * b + c * c + d * d + e * e);
}
bool check(int i, int j, int k) {
  long double l = (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]);
  long double m =
      len(a[j] - a[i], b[j] - b[i], c[j] - c[i], d[j] - d[i], e[j] - e[i]) *
      len(a[k] - a[i], b[k] - b[i], c[k] - c[i], d[k] - d[i], e[k] - e[i]);
  return acos(l / m) < pi / 2;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  srand(time(NULL));
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i] >> b[i] >> c[i] >> d[i] >> e[i];
    p.push_back(i);
  }
  random_shuffle((p).begin(), (p).end());
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      for (int k = 0; k < n; ++k) {
        if (i != j && i != k && j != k && check(p[i], p[j], p[k]))
          zly[p[i]] = 1;
        if (zly[p[i]]) break;
      }
      if (zly[p[i]]) break;
    }
  }
  cout << n - zly.count() << '\n';
  for (int i = 1; i <= n; ++i) {
    if (!zly[i]) cout << i << '\n';
  }
}

### Prompt

Create a solution in cpp for 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 = 1007;
const long double pi = acos(-1);
const long double eps = 1e-9;
int n;
int a[N];
int b[N];
int c[N];
int d[N];
int e[N];
vector<int> p;
bitset<N> zly;
long double len(long double a, long double b, long double c, long double d,
                long double e) {
  return sqrt(a * a + b * b + c * c + d * d + e * e);
}
bool check(int i, int j, int k) {
  long double l = (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]);
  long double m =
      len(a[j] - a[i], b[j] - b[i], c[j] - c[i], d[j] - d[i], e[j] - e[i]) *
      len(a[k] - a[i], b[k] - b[i], c[k] - c[i], d[k] - d[i], e[k] - e[i]);
  return acos(l / m) < pi / 2;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  srand(time(NULL));
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i] >> b[i] >> c[i] >> d[i] >> e[i];
    p.push_back(i);
  }
  random_shuffle((p).begin(), (p).end());
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      for (int k = 0; k < n; ++k) {
        if (i != j && i != k && j != k && check(p[i], p[j], p[k]))
          zly[p[i]] = 1;
        if (zly[p[i]]) break;
      }
      if (zly[p[i]]) break;
    }
  }
  cout << n - zly.count() << '\n';
  for (int i = 1; i <= n; ++i) {
    if (!zly[i]) cout << i << '\n';
  }
}

```