#include <bits/stdc++.h>
using namespace std;
int prod(vector<int> A, vector<int> B) {
  int ans = 0;
  for (int i = 0; i < A.size(); i++) ans += A[i] * B[i];
  return ans;
}
int main() {
  int n;
  cin >> n;
  if (n > 100) {
    cout << "0\n";
    return 0;
  }
  vector<vector<int> > A(n, vector<int>(5));
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 5; j++) cin >> A[i][j];
  vector<int> ans;
  for (int a = 0; a < n; a++) {
    int ok = true;
    for (int b = 0; b < n && ok; b++)
      if (b != a)
        for (int c = b + 1; c < n && ok; c++)
          if (c != a) {
            vector<int> x(5), y(5);
            for (int j = 0; j < 5; j++) {
              x[j] = A[b][j] - A[a][j];
              y[j] = A[c][j] - A[a][j];
            }
            int xy = prod(x, y);
            if (xy > 0) {
              ok = false;
            }
          }
    if (ok) ans.push_back(a + 1);
  }
  cout << ans.size() << endl;
  for (int x : ans) cout << x << endl;
  return 0;
}

### Prompt

Please formulate 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 namespace std;
int prod(vector<int> A, vector<int> B) {
  int ans = 0;
  for (int i = 0; i < A.size(); i++) ans += A[i] * B[i];
  return ans;
}
int main() {
  int n;
  cin >> n;
  if (n > 100) {
    cout << "0\n";
    return 0;
  }
  vector<vector<int> > A(n, vector<int>(5));
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 5; j++) cin >> A[i][j];
  vector<int> ans;
  for (int a = 0; a < n; a++) {
    int ok = true;
    for (int b = 0; b < n && ok; b++)
      if (b != a)
        for (int c = b + 1; c < n && ok; c++)
          if (c != a) {
            vector<int> x(5), y(5);
            for (int j = 0; j < 5; j++) {
              x[j] = A[b][j] - A[a][j];
              y[j] = A[c][j] - A[a][j];
            }
            int xy = prod(x, y);
            if (xy > 0) {
              ok = false;
            }
          }
    if (ok) ans.push_back(a + 1);
  }
  cout << ans.size() << endl;
  for (int x : ans) cout << x << endl;
  return 0;
}

```