#include <bits/stdc++.h>
using namespace std;
struct pt {
  int a, b, c, d, e;
};
double res(pt x, pt y) {
  double ans = x.a * y.a + x.b * y.b + x.c * y.c + x.d * y.d + x.e * y.e;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  pt z[n];
  int good[n];
  for (int i = 0; i < n; i++) {
    cin >> z[i].a >> z[i].b >> z[i].c >> z[i].d >> z[i].e;
    good[i] = 1;
  }
  for (int i = 0; i < n; i++) {
    if (!good[i]) continue;
    bool valid = true;
    for (int x = 0; x < n; x++) {
      if (x == i) continue;
      pt l1;
      l1.a = z[i].a - z[x].a;
      l1.b = z[i].b - z[x].b;
      l1.c = z[i].c - z[x].c;
      l1.d = z[i].d - z[x].d;
      l1.e = z[i].e - z[x].e;
      for (int y = 0; y < n; y++) {
        if (y == i || y == x) continue;
        pt l2;
        l2.a = z[i].a - z[y].a;
        l2.b = z[i].b - z[y].b;
        l2.c = z[i].c - z[y].c;
        l2.d = z[i].d - z[y].d;
        l2.e = z[i].e - z[y].e;
        if (l1.a * l2.a + l1.b * l2.b + l1.c * l2.c + l1.d * l2.d +
                l1.e * l2.e >
            0) {
          valid = false;
          good[i] = false;
          break;
        }
      }
      if (!valid) break;
    }
  }
  vector<long long int> ans;
  for (int i = 0; i < n; i++) {
    if (good[i]) {
      ans.push_back(i + 1);
    }
  }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) cout << ans[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;
struct pt {
  int a, b, c, d, e;
};
double res(pt x, pt y) {
  double ans = x.a * y.a + x.b * y.b + x.c * y.c + x.d * y.d + x.e * y.e;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  pt z[n];
  int good[n];
  for (int i = 0; i < n; i++) {
    cin >> z[i].a >> z[i].b >> z[i].c >> z[i].d >> z[i].e;
    good[i] = 1;
  }
  for (int i = 0; i < n; i++) {
    if (!good[i]) continue;
    bool valid = true;
    for (int x = 0; x < n; x++) {
      if (x == i) continue;
      pt l1;
      l1.a = z[i].a - z[x].a;
      l1.b = z[i].b - z[x].b;
      l1.c = z[i].c - z[x].c;
      l1.d = z[i].d - z[x].d;
      l1.e = z[i].e - z[x].e;
      for (int y = 0; y < n; y++) {
        if (y == i || y == x) continue;
        pt l2;
        l2.a = z[i].a - z[y].a;
        l2.b = z[i].b - z[y].b;
        l2.c = z[i].c - z[y].c;
        l2.d = z[i].d - z[y].d;
        l2.e = z[i].e - z[y].e;
        if (l1.a * l2.a + l1.b * l2.b + l1.c * l2.c + l1.d * l2.d +
                l1.e * l2.e >
            0) {
          valid = false;
          good[i] = false;
          break;
        }
      }
      if (!valid) break;
    }
  }
  vector<long long int> ans;
  for (int i = 0; i < n; i++) {
    if (good[i]) {
      ans.push_back(i + 1);
    }
  }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << endl;
  return 0;
}

```