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;
int Ma[2000][5];
double calc(int x[], int y[], int z[]) {
  double res = 0;
  for (int i = 0; i < 5; ++i) {
    res += (y[i] - x[i]) * (z[i] - x[i]);
  }
  return res;
}
int main() {
  int n;
  vector<int> ans;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 5; ++j) {
      cin >> Ma[i][j];
    }
  }
  for (int i = 0; i < n; ++i) {
    int found = 0;
    for (int j = 0; j < n; ++j) {
      if (i == j) {
        continue;
      }
      for (int k = 0; k < n; ++k) {
        if ((i == j) || (j == k) || (k == i)) {
          continue;
        }
        if (acos((calc(Ma[i], Ma[j], Ma[k]) /
                  ((sqrt(calc(Ma[i], Ma[j], Ma[j]))) *
                   (sqrt(calc(Ma[i], Ma[k], Ma[k])))))) < acos(0)) {
          found = 1;
          break;
        }
      }
      if (found) {
        break;
      }
    }
    if (!found) {
      ans.push_back(i + 1);
    }
  }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); ++i) {
    cout << ans[i] << endl;
  }
  return 0;
}
