#include <bits/stdc++.h>
using namespace std;
struct vect {
  long double x[5];
  void get() {
    for (int i = 0; i < 5; i++) cin >> x[i];
  }
};
vect operator-(vect a, vect b) {
  vect c;
  for (int i = 0; i < 5; i++) c.x[i] = a.x[i] - b.x[i];
  return c;
}
vect operator^(vect a, vect b) { return b - a; }
long double operator*(vect a, vect b) {
  long double ret = 0;
  for (int i = 0; i < 5; i++) ret += a.x[i] * b.x[i];
  return ret;
}
long double mod(vect a) { return sqrtl(a * a); }
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int n;
  cin >> n;
  vect A[n];
  for (int i = 0; i < n; i++) A[i].get();
  vector<int> ats;
  for (int i = 0; i < n; i++) {
    bool good = true;
    for (int a = 0; a < n && good; a++) {
      if (a == i) continue;
      for (int b = 0; b < n && good; b++) {
        if (b == i || b == a) continue;
        vect x = A[i] ^ A[b];
        vect y = A[i] ^ A[a];
        if (acosl((x * y) / (mod(x) * mod(y))) + (1e-14) <
            (3.14159265358979323846264 / 2.0)) {
          good = false;
        }
      }
    }
    if (good) ats.push_back(i);
  }
  cout << ats.size() << "\n";
  for (int i : ats) cout << i + 1 << " ";
}

### 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>
using namespace std;
struct vect {
  long double x[5];
  void get() {
    for (int i = 0; i < 5; i++) cin >> x[i];
  }
};
vect operator-(vect a, vect b) {
  vect c;
  for (int i = 0; i < 5; i++) c.x[i] = a.x[i] - b.x[i];
  return c;
}
vect operator^(vect a, vect b) { return b - a; }
long double operator*(vect a, vect b) {
  long double ret = 0;
  for (int i = 0; i < 5; i++) ret += a.x[i] * b.x[i];
  return ret;
}
long double mod(vect a) { return sqrtl(a * a); }
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int n;
  cin >> n;
  vect A[n];
  for (int i = 0; i < n; i++) A[i].get();
  vector<int> ats;
  for (int i = 0; i < n; i++) {
    bool good = true;
    for (int a = 0; a < n && good; a++) {
      if (a == i) continue;
      for (int b = 0; b < n && good; b++) {
        if (b == i || b == a) continue;
        vect x = A[i] ^ A[b];
        vect y = A[i] ^ A[a];
        if (acosl((x * y) / (mod(x) * mod(y))) + (1e-14) <
            (3.14159265358979323846264 / 2.0)) {
          good = false;
        }
      }
    }
    if (good) ats.push_back(i);
  }
  cout << ats.size() << "\n";
  for (int i : ats) cout << i + 1 << " ";
}

```