#include <bits/stdc++.h>
using namespace std;
struct vector5 {
  int c1, c2, c3, c4, c5;
  double dot(const vector5& other) const {
    return c1 * other.c1 + c2 * other.c2 + c3 * other.c3 + c4 * other.c4 +
           c5 * other.c5;
  }
  double len() const { return sqrt(this->dot(*this)); }
};
vector5 operator-(vector5& l, vector5& r) {
  return {l.c1 - r.c1, l.c2 - r.c2, l.c3 - r.c3, l.c4 - r.c4, l.c5 - r.c5};
  ;
}
istream& operator>>(istream& is, vector5& v) {
  is >> v.c1 >> v.c2 >> v.c3 >> v.c4 >> v.c5;
  return is;
}
int n, cntBad;
vector5 v[1005];
bool bad[1005];
double angle(const vector5& a, const vector5& b) {
  return acos(a.dot(b) / a.len() / b.len()) / M_PI * 180;
}
bool isBad(int i, int j, int k) { return angle(v[j] - v[i], v[k] - v[i]) < 90; }
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
  }
  for (int i = 1; i <= n - 2; i++) {
    for (int j = i + 1; j <= n - 1; j++) {
      for (int k = j + 1; k <= n; k++) {
        if (!bad[i] && isBad(i, j, k)) {
          cntBad++;
          bad[i] = true;
        }
        if (!bad[j] && isBad(j, i, k)) {
          cntBad++;
          bad[j] = true;
        }
        if (!bad[k] && isBad(k, j, i)) {
          cntBad++;
          bad[k] = true;
        }
      }
    }
  }
  cout << n - cntBad << '\n';
  for (int i = 1; i <= n; i++)
    if (!bad[i]) cout << i << ' ';
}

### 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;
struct vector5 {
  int c1, c2, c3, c4, c5;
  double dot(const vector5& other) const {
    return c1 * other.c1 + c2 * other.c2 + c3 * other.c3 + c4 * other.c4 +
           c5 * other.c5;
  }
  double len() const { return sqrt(this->dot(*this)); }
};
vector5 operator-(vector5& l, vector5& r) {
  return {l.c1 - r.c1, l.c2 - r.c2, l.c3 - r.c3, l.c4 - r.c4, l.c5 - r.c5};
  ;
}
istream& operator>>(istream& is, vector5& v) {
  is >> v.c1 >> v.c2 >> v.c3 >> v.c4 >> v.c5;
  return is;
}
int n, cntBad;
vector5 v[1005];
bool bad[1005];
double angle(const vector5& a, const vector5& b) {
  return acos(a.dot(b) / a.len() / b.len()) / M_PI * 180;
}
bool isBad(int i, int j, int k) { return angle(v[j] - v[i], v[k] - v[i]) < 90; }
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
  }
  for (int i = 1; i <= n - 2; i++) {
    for (int j = i + 1; j <= n - 1; j++) {
      for (int k = j + 1; k <= n; k++) {
        if (!bad[i] && isBad(i, j, k)) {
          cntBad++;
          bad[i] = true;
        }
        if (!bad[j] && isBad(j, i, k)) {
          cntBad++;
          bad[j] = true;
        }
        if (!bad[k] && isBad(k, j, i)) {
          cntBad++;
          bad[k] = true;
        }
      }
    }
  }
  cout << n - cntBad << '\n';
  for (int i = 1; i <= n; i++)
    if (!bad[i]) cout << i << ' ';
}

```