#include <bits/stdc++.h>
using namespace std;
struct point {
  vector<long double> x;
  point() : x(5) {}
  point(long double q, long double w, long double e, long double r,
        long double t)
      : x({q, w, e, r, t}) {}
  long double operator*(const point& b) const {
    long double ret = 0.0;
    for (int i = 0; i < 5; i++) ret += this->x[i] * b.x[i];
    return ret;
  }
  point operator-(const point& b) const {
    point ret;
    for (int i = 0; i < 5; i++) ret.x[i] = this->x[i] - b.x[i];
    return ret;
  }
  long double len() const { return sqrt((*this) * (*this)); }
};
long double angle(const point& a, const point& b) {
  return (a * b) / (a.len() * b.len());
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<point> points(n);
  for (auto& e : points) {
    for (auto& k : e.x) cin >> k;
  }
  if (n > 33) return cout << 0, 0;
  if (n <= 2) {
    cout << n << '\n';
    for (int i = 1; i <= n; i++) cout << i << '\n';
    return 0;
  }
  for (int x = 0; x < n; x++) {
    bool f = 1;
    for (int a = 0; a < n; a++) {
      if (a == x) continue;
      for (int b = 0; b < n; b++) {
        if (b == x || b == a) continue;
        point xa = points[a] - points[x];
        point xb = points[b] - points[x];
        if (angle(xa, xb) > 0) f = 0;
      }
    }
    if (f) return cout << 1 << '\n' << x + 1, 0;
  }
  cout << 0;
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 point {
  vector<long double> x;
  point() : x(5) {}
  point(long double q, long double w, long double e, long double r,
        long double t)
      : x({q, w, e, r, t}) {}
  long double operator*(const point& b) const {
    long double ret = 0.0;
    for (int i = 0; i < 5; i++) ret += this->x[i] * b.x[i];
    return ret;
  }
  point operator-(const point& b) const {
    point ret;
    for (int i = 0; i < 5; i++) ret.x[i] = this->x[i] - b.x[i];
    return ret;
  }
  long double len() const { return sqrt((*this) * (*this)); }
};
long double angle(const point& a, const point& b) {
  return (a * b) / (a.len() * b.len());
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<point> points(n);
  for (auto& e : points) {
    for (auto& k : e.x) cin >> k;
  }
  if (n > 33) return cout << 0, 0;
  if (n <= 2) {
    cout << n << '\n';
    for (int i = 1; i <= n; i++) cout << i << '\n';
    return 0;
  }
  for (int x = 0; x < n; x++) {
    bool f = 1;
    for (int a = 0; a < n; a++) {
      if (a == x) continue;
      for (int b = 0; b < n; b++) {
        if (b == x || b == a) continue;
        point xa = points[a] - points[x];
        point xb = points[b] - points[x];
        if (angle(xa, xb) > 0) f = 0;
      }
    }
    if (f) return cout << 1 << '\n' << x + 1, 0;
  }
  cout << 0;
  return 0;
}

```