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;
const int maxn = 1e3 + 10;
struct PT {
  long long a, b, c, d, e;
  PT() {}
  PT(long long a, long long b, long long c, long long d, long long e)
      : a(a), b(b), c(c), d(d), e(e) {}
  PT operator-(const PT &p) {
    return PT(a - p.a, b - p.b, c - p.c, d - p.d, e - p.e);
  }
};
long long dot(PT p, PT q) {
  return p.a * q.a + p.b * q.b + p.c * q.c + p.d * q.d + p.e * q.e;
}
int readInt() {
  int tmp;
  cin >> tmp;
  return tmp;
}
PT pts[maxn];
int n;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int(i) = (0); (i) <= (n - 1); (i)++) {
    cin >> pts[i].a >> pts[i].b >> pts[i].c >> pts[i].d >> pts[i].e;
  }
  if (n > 200) {
    cout << 0 << endl;
    return 0;
  }
  set<int> bad;
  for (int(a) = (0); (a) <= (n - 1); (a)++) {
    for (int(b) = (0); (b) <= (n - 1); (b)++) {
      for (int(c) = (0); (c) <= (n - 1); (c)++) {
        if (a == b || a == c || b == c) continue;
        if (dot(pts[b] - pts[a], pts[c] - pts[a]) > 0) {
          bad.insert(a);
        }
      }
    }
  }
  cout << n - bad.size() << endl;
  for (int(i) = (0); (i) <= (n - 1); (i)++) {
    if (!bad.count(i)) cout << i + 1 << endl;
  }
}
