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;
long long gcd(long long x, long long y) {
  if (y == 0)
    return x;
  else
    return gcd(y, x % y);
}
long long expo(long long n, long long m) {
  long long r = 1;
  while (m > 0) {
    if (m % 2) r = (r * n) % 1000000007;
    n = (n * n) % 1000000007;
    m = m / 2;
  }
  return r % 1000000007;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  vector<long long> adj[100000];
  for (int i = 0; i < n; i++) {
    long long a, b, c, d, e;
    cin >> a >> b >> c >> d >> e;
    adj[i].push_back(a), adj[i].push_back(b), adj[i].push_back(c),
        adj[i].push_back(d), adj[i].push_back(e);
  }
  set<long long> ans;
  for (long long i = 0; i < n; i++) {
    vector<long long> diff[10000];
    long long d = 0;
    for (long long j = 0; j < n; j++) {
      if (i == j) continue;
      for (long long k = 0; k < 5; k++)
        diff[d].push_back(adj[j][k] - adj[i][k]);
      d++;
    }
    bool ok = false;
    for (long long j = 0; j < d; j++) {
      for (long long k = j + 1; k < d; k++) {
        long long an = 0;
        for (long long m = 0; m < 5; m++) {
          an = an + diff[j][m] * diff[k][m];
        }
        if (an > 0) {
          ok = true;
          break;
        }
      }
      if (ok) break;
    }
    if (!ok) ans.insert(i);
  }
  cout << ans.size() << '\n';
  set<long long>::iterator it = ans.begin();
  for (; it != ans.end(); it++) cout << *it + 1 << '\n';
  return 0;
}
