#include <bits/stdc++.h>
#pragma GCC optimize("Ofast", "unroll-loops")
using namespace std;
const long long inf = (1ll << 60);
const double PI = acos(-1);
const int iinf = 2147483647;
const long long mod = 1e9 + 7;
const long long maxn = 1e4 + 5;
struct Point {
  double a, b, c, d, e;
};
double dot(Point A, Point B) {
  return A.a * B.a + A.b * B.b + A.c * B.c + A.d * B.d + A.e * B.e;
}
Point sub(Point A, Point B) {
  return {A.a - B.a, A.b - B.b, A.c - B.c, A.d - B.d, A.e - B.e};
}
signed main() {
  int n;
  cin >> n;
  vector<Point> vc(n);
  for (int i = (0); i < (n); i++)
    cin >> vc[i].a >> vc[i].b >> vc[i].c >> vc[i].d >> vc[i].e;
  if (n > 11) {
    cout << 0 << endl;
    return 0;
  }
  int cnt = 0;
  vector<int> ans;
  for (int i = (0); i < (n); i++) {
    bool die = false;
    for (int j = (0); j < (n); j++) {
      if (j == i) continue;
      if (die) break;
      for (int k = (0); k < (n); k++) {
        if (k == j || k == i) continue;
        double top = dot(sub(vc[j], vc[i]), sub(vc[k], vc[i]));
        double bot = sqrt(dot(sub(vc[j], vc[i]), sub(vc[j], vc[i]))) *
                     sqrt(dot(sub(vc[k], vc[i]), sub(vc[k], vc[i])));
        double res = acos(top / bot);
        if (res < PI / 2) {
          die = true;
          break;
        }
      }
    }
    if (!die) {
      cnt++;
      ans.push_back(i + 1);
    }
  }
  cout << cnt << endl;
  for (int i = (0); i < (cnt); i++) cout << ans[i] << endl;
}

### 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>
#pragma GCC optimize("Ofast", "unroll-loops")
using namespace std;
const long long inf = (1ll << 60);
const double PI = acos(-1);
const int iinf = 2147483647;
const long long mod = 1e9 + 7;
const long long maxn = 1e4 + 5;
struct Point {
  double a, b, c, d, e;
};
double dot(Point A, Point B) {
  return A.a * B.a + A.b * B.b + A.c * B.c + A.d * B.d + A.e * B.e;
}
Point sub(Point A, Point B) {
  return {A.a - B.a, A.b - B.b, A.c - B.c, A.d - B.d, A.e - B.e};
}
signed main() {
  int n;
  cin >> n;
  vector<Point> vc(n);
  for (int i = (0); i < (n); i++)
    cin >> vc[i].a >> vc[i].b >> vc[i].c >> vc[i].d >> vc[i].e;
  if (n > 11) {
    cout << 0 << endl;
    return 0;
  }
  int cnt = 0;
  vector<int> ans;
  for (int i = (0); i < (n); i++) {
    bool die = false;
    for (int j = (0); j < (n); j++) {
      if (j == i) continue;
      if (die) break;
      for (int k = (0); k < (n); k++) {
        if (k == j || k == i) continue;
        double top = dot(sub(vc[j], vc[i]), sub(vc[k], vc[i]));
        double bot = sqrt(dot(sub(vc[j], vc[i]), sub(vc[j], vc[i]))) *
                     sqrt(dot(sub(vc[k], vc[i]), sub(vc[k], vc[i])));
        double res = acos(top / bot);
        if (res < PI / 2) {
          die = true;
          break;
        }
      }
    }
    if (!die) {
      cnt++;
      ans.push_back(i + 1);
    }
  }
  cout << cnt << endl;
  for (int i = (0); i < (cnt); i++) cout << ans[i] << endl;
}

```