#include <bits/stdc++.h>
using ll = long long;
using ld = long double;
using namespace std;
class A {
 public:
  int a, b, c, d, e;
  A(int a, int b, int c, int d, int e) : a(a), b(b), c(c), d(d), e(e) {}
};
double scalar(A a, A b) {
  double result = 0.0;
  result += a.a * b.a;
  result += a.b * b.b;
  result += a.c * b.c;
  result += a.d * b.d;
  result += a.e * b.e;
  return result;
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  if (n < 3) {
    cout << n << endl;
    for (int i = 0; i < n; ++i) cout << i + 1 << endl;
    return 0;
  }
  vector<A> points;
  for (int i = 0; i < n; ++i) {
    int a, b, c, d, e;
    cin >> a >> b >> c >> d >> e;
    points.push_back(A(a, b, c, d, e));
  }
  int count = 0;
  vector<int> L;
  for (int i = 0; i < n; ++i) {
    A X = points[i];
    bool skip = false;
    for (int j = 0; j < n && !skip; ++j) {
      if (i == j) continue;
      A Y = points[j];
      Y.a -= X.a;
      Y.b -= X.b;
      Y.c -= X.c;
      Y.d -= X.d;
      Y.e -= X.e;
      for (int k = 0; k < n && !skip; ++k) {
        if ((k == j) || (k == i)) continue;
        A Z = points[k];
        Z.a -= X.a;
        Z.b -= X.b;
        Z.c -= X.c;
        Z.d -= X.d;
        Z.e -= X.e;
        double value =
            scalar(Y, Z) / pow(scalar(Y, Y), 0.5) / pow(scalar(Z, Z), 0.5);
        if (value > 0.0) {
          skip = true;
        }
      }
    }
    if (!skip) {
      count++;
      L.push_back(i + 1);
    }
  }
  cout << count << endl;
  for (int i = 0; i < L.size(); ++i) cout << L[i] << endl;
  return 0;
}

### Prompt

Your challenge is to write a cpp solution 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 ll = long long;
using ld = long double;
using namespace std;
class A {
 public:
  int a, b, c, d, e;
  A(int a, int b, int c, int d, int e) : a(a), b(b), c(c), d(d), e(e) {}
};
double scalar(A a, A b) {
  double result = 0.0;
  result += a.a * b.a;
  result += a.b * b.b;
  result += a.c * b.c;
  result += a.d * b.d;
  result += a.e * b.e;
  return result;
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  if (n < 3) {
    cout << n << endl;
    for (int i = 0; i < n; ++i) cout << i + 1 << endl;
    return 0;
  }
  vector<A> points;
  for (int i = 0; i < n; ++i) {
    int a, b, c, d, e;
    cin >> a >> b >> c >> d >> e;
    points.push_back(A(a, b, c, d, e));
  }
  int count = 0;
  vector<int> L;
  for (int i = 0; i < n; ++i) {
    A X = points[i];
    bool skip = false;
    for (int j = 0; j < n && !skip; ++j) {
      if (i == j) continue;
      A Y = points[j];
      Y.a -= X.a;
      Y.b -= X.b;
      Y.c -= X.c;
      Y.d -= X.d;
      Y.e -= X.e;
      for (int k = 0; k < n && !skip; ++k) {
        if ((k == j) || (k == i)) continue;
        A Z = points[k];
        Z.a -= X.a;
        Z.b -= X.b;
        Z.c -= X.c;
        Z.d -= X.d;
        Z.e -= X.e;
        double value =
            scalar(Y, Z) / pow(scalar(Y, Y), 0.5) / pow(scalar(Z, Z), 0.5);
        if (value > 0.0) {
          skip = true;
        }
      }
    }
    if (!skip) {
      count++;
      L.push_back(i + 1);
    }
  }
  cout << count << endl;
  for (int i = 0; i < L.size(); ++i) cout << L[i] << endl;
  return 0;
}

```