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 Ans[1001], A[1001], B[1001], C[1001], D[1001], E[1001];
int value(int Arr[], int i, int j, int k) {
  return (Arr[j] - Arr[i]) * (Arr[k] - Arr[i]);
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> A[i] >> B[i] >> C[i] >> D[i] >> E[i];
  }
  int index = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (j != i) {
        for (int k = 0; k < n; k++) {
          if (k != i && k != j) {
            if ((value(A, i, j, k) + value(B, i, j, k) + value(C, i, j, k) +
                 value(D, i, j, k) + value(E, i, j, k)) > 0) {
              Ans[i] = -1;
              break;
            }
          }
        }
      }
    }
  }
  int count = 0;
  for (int i = 0; i < n; i++) {
    if (Ans[i] != -1) {
      count++;
    }
  }
  cout << count << endl;
  for (int i = 0; i < n; i++) {
    if (Ans[i] != -1) {
      cout << i + 1 << endl;
      count++;
    }
  }
  return 0;
}
