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 p[1005][6];
vector<int> ans;
int cal(int i, int j, int k) {
  int ans = 0;
  for (int q = 1; q <= 5; q++) {
    int x = p[j][q] - p[i][q];
    int y = p[k][q] - p[i][q];
    ans += x * y;
  }
  return ans;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p[i][1] >> p[i][2] >> p[i][3] >> p[i][4] >> p[i][5];
  }
  for (int i = 1; i <= n; i++) {
    int tag = 1;
    for (int j = 1; j <= n; j++) {
      if (j == i) continue;
      for (int k = 1; k <= n; k++) {
        if (k == j || k == i) continue;
        if (cal(i, j, k) > 0) {
          tag = 0;
          break;
        }
      }
      if (!tag) break;
    }
    if (tag) ans.push_back(i);
  }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i] << endl;
  }
}
