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;
struct fd {
  long long int x1, x2, x3, x4, x5;
};
fd a[1000005];
long long int pro(fd a, fd b) {
  long long int ans = 0;
  ans += a.x1 * b.x1;
  ans += a.x2 * b.x2;
  ans += a.x3 * b.x3;
  ans += a.x4 * b.x4;
  ans += a.x5 * b.x5;
  return ans;
}
long long int ans[1000005];
int main() {
  long long int n;
  scanf("%lld", &n);
  for (long long int i = 1; i <= n; i++)
    cin >> a[i].x1 >> a[i].x2 >> a[i].x3 >> a[i].x4 >> a[i].x5;
  long long int tot = n;
  for (long long int i = 1; i <= n; i++) {
    ans[i] = 1;
    for (long long int j = 1; j <= n; j++) {
      if (ans[i] == 0) break;
      if (i == j) continue;
      for (long long int k = 1; k <= n; k++) {
        if (k == i || k == j) continue;
        fd f1;
        f1.x1 = a[k].x1 - a[i].x1;
        f1.x2 = a[k].x2 - a[i].x2;
        f1.x3 = a[k].x3 - a[i].x3;
        f1.x4 = a[k].x4 - a[i].x4;
        f1.x5 = a[k].x5 - a[i].x5;
        fd f2;
        f2.x1 = a[j].x1 - a[i].x1;
        f2.x2 = a[j].x2 - a[i].x2;
        f2.x3 = a[j].x3 - a[i].x3;
        f2.x4 = a[j].x4 - a[i].x4;
        f2.x5 = a[j].x5 - a[i].x5;
        if (pro(f1, f2) > 0) {
          ans[i] = 0;
          tot--;
          break;
        }
      }
    }
  }
  printf("%lld\n", tot);
  for (long long int i = 1; i <= n; i++) {
    if (ans[i]) printf("%lld ", i);
  }
  return 0;
}
