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;
long long mod = 1000000007;
int main() {
  std::ios::sync_with_stdio(false);
  long long n, k = 0;
  cin >> n;
  vector<vector<int> > a(n, vector<int>(5, 0));
  for (int i = (int)(0); i < ((int)n); i++) {
    cin >> a[i][0] >> a[i][1] >> a[i][2] >> a[i][3] >> a[i][4];
  }
  vector<int> good(n, 1);
  for (int i = (int)(0); i < ((int)n); i++) {
    bool notGood = 0;
    for (int j = (int)(0); j < ((int)n); j++) {
      if (i == j) continue;
      for (int k = (int)(j + 1); k < ((int)n); k++) {
        if (k == i) continue;
        long long sign = 0;
        for (int p = (int)(0); p < ((int)5); p++)
          sign += (a[j][p] - a[i][p]) * (a[k][p] - a[i][p]);
        if (sign > 0) {
          good[i] = 0;
          notGood = 1;
          break;
        }
        if (notGood) break;
      }
      if (notGood) break;
    }
  }
  for (int i = (int)(0); i < ((int)n); i++) {
    k += good[i];
  }
  cout << k << endl;
  for (int i = (int)(0); i < ((int)n); i++) {
    if (good[i]) cout << i + 1 << endl;
  }
  return 0;
}
