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 arr[1007][5];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) cin >> arr[i][j];
  }
  vector<int> v;
  for (int i = 0; i < n; i++) {
    int flag = 0;
    for (int j = 0; j < n; j++) {
      for (int h = j + 1; h < n; h++) {
        if (j == i && h == i) continue;
        int val = 0;
        for (int k = 0; k < 5; k++)
          val += (arr[j][k] - arr[i][k]) * (arr[h][k] - arr[i][k]);
        if (val > 0) {
          flag = 1;
          break;
        }
      }
      if (flag == 1) break;
    }
    if (flag == 0) v.push_back(i + 1);
  }
  int jn = v.size();
  cout << jn << endl;
  for (int i = 0; i < jn; i++) cout << v[i] << endl;
}
