#include <bits/stdc++.h>
using namespace std;
const double eps = 0.0000001;
struct fun {
  double a, b, c, d, e;
};
int n;
vector<int> v;
fun m[500];
double dist(double a, double b, double c, double d, double e) {
  return sqrt(a * a + b * b + c * c + d * d + e * e);
}
int main() {
  ios_base::sync_with_stdio(0);
  int i, j, k;
  double half = M_PI / 2;
  cin >> n;
  if (n > 300) {
    cout << 0;
    return 0;
  }
  for (i = 1; i <= n; i++)
    cin >> m[i].a >> m[i].b >> m[i].c >> m[i].d >> m[i].e;
  for (i = 1; i <= n; i++) {
    double u = 213;
    for (j = 1; j <= n; j++)
      if (j != i)
        for (k = 1; k <= n; k++)
          if (k != j && k != i) {
            fun x = {m[i].a - m[j].a, m[i].b - m[j].b, m[i].c - m[j].c,
                     m[i].d - m[j].d, m[i].e - m[j].e};
            fun y = {m[i].a - m[k].a, m[i].b - m[k].b, m[i].c - m[k].c,
                     m[i].d - m[k].d, m[i].e - m[k].e};
            double l1 = dist(m[i].a - m[j].a, m[i].b - m[j].b, m[i].c - m[j].c,
                             m[i].d - m[j].d, m[i].e - m[j].e);
            double l2 = dist(m[i].a - m[k].a, m[i].b - m[k].b, m[i].c - m[k].c,
                             m[i].d - m[k].d, m[i].e - m[k].e);
            double p =
                x.a * y.a + x.b * y.b + x.c * y.c + x.d * y.d + x.e * y.e;
            double curr = acos(p / (l1 * l2));
            u = min(u, curr);
          }
    if (u < M_PI / 2) continue;
    v.push_back(i);
  }
  cout << v.size() << endl;
  for (i = 0; i < v.size(); i++) cout << v[i] << endl;
  return 0;
}

### Prompt

Please create a solution in cpp 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 double eps = 0.0000001;
struct fun {
  double a, b, c, d, e;
};
int n;
vector<int> v;
fun m[500];
double dist(double a, double b, double c, double d, double e) {
  return sqrt(a * a + b * b + c * c + d * d + e * e);
}
int main() {
  ios_base::sync_with_stdio(0);
  int i, j, k;
  double half = M_PI / 2;
  cin >> n;
  if (n > 300) {
    cout << 0;
    return 0;
  }
  for (i = 1; i <= n; i++)
    cin >> m[i].a >> m[i].b >> m[i].c >> m[i].d >> m[i].e;
  for (i = 1; i <= n; i++) {
    double u = 213;
    for (j = 1; j <= n; j++)
      if (j != i)
        for (k = 1; k <= n; k++)
          if (k != j && k != i) {
            fun x = {m[i].a - m[j].a, m[i].b - m[j].b, m[i].c - m[j].c,
                     m[i].d - m[j].d, m[i].e - m[j].e};
            fun y = {m[i].a - m[k].a, m[i].b - m[k].b, m[i].c - m[k].c,
                     m[i].d - m[k].d, m[i].e - m[k].e};
            double l1 = dist(m[i].a - m[j].a, m[i].b - m[j].b, m[i].c - m[j].c,
                             m[i].d - m[j].d, m[i].e - m[j].e);
            double l2 = dist(m[i].a - m[k].a, m[i].b - m[k].b, m[i].c - m[k].c,
                             m[i].d - m[k].d, m[i].e - m[k].e);
            double p =
                x.a * y.a + x.b * y.b + x.c * y.c + x.d * y.d + x.e * y.e;
            double curr = acos(p / (l1 * l2));
            u = min(u, curr);
          }
    if (u < M_PI / 2) continue;
    v.push_back(i);
  }
  cout << v.size() << endl;
  for (i = 0; i < v.size(); i++) cout << v[i] << endl;
  return 0;
}

```