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.
#include <bits/stdc++.h>
using namespace std;
double ls[1003][5];
vector<int> ans;
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      cin >> ls[i][j];
    }
  }
  if (n > 100)
    cout << 0 << endl;
  else {
    for (int i = 0; i < n; i++) {
      bool check = 1;
      for (int j = 0; j < n; j++) {
        for (int k = j + 1; k < n; k++) {
          if (j == i || k == i) {
            continue;
          }
          double s = 0;
          for (int z = 0; z < 5; z++) {
            s += (ls[j][z] - ls[i][z]) * (ls[k][z] - ls[i][z]);
          }
          if (s > 0) {
            check = 0;
          }
        }
      }
      if (check) ans.push_back(i + 1);
    }
    cout << ans.size() << endl;
    if (ans.size() > 0) {
      cout << ans[0];
      for (int i = 1; i < ans.size(); i++) {
        cout << " " << ans[i];
      }
      cout << endl;
    }
  }
}
