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 long long INF = 1e15;
const int MOD = 1e9 + 7;
const double EPS = 1e-13;
const double PI = acos(-1.0);
mt19937 rng((int)chrono::steady_clock::now().time_since_epoch().count());
const int N = 1005;
long long n, x[N][5];
long long dot(long long a[], long long b[], long long c[]) {
  long long ret = 0;
  for (int i = 0; i < 5; ++i) ret += (b[i] - a[i]) * (c[i] - a[i]);
  return ret;
}
double dist(long long a[], long long b[]) {
  double ret = 0;
  for (int i = 0; i < 5; ++i) ret += (b[i] - a[i]) * (b[i] - a[i]);
  return sqrt(ret);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < 5; ++j) cin >> x[i][j];
  vector<int> ans;
  for (int i = 0; i < n; ++i) {
    bool ok = true;
    for (int j = 0; j < n; ++j) {
      if (j == i) continue;
      for (int k = j + 1; k < n; ++k) {
        if (k == i) continue;
        if (acos(dot(x[i], x[j], x[k]) /
                 (dist(x[i], x[j]) * dist(x[i], x[k]))) +
                EPS <=
            PI / 2.0) {
          ok = false;
          break;
        }
      }
      if (!ok) break;
    }
    if (ok) ans.push_back(i);
  }
  cout << (int)ans.size() << "\n";
  for (auto x : ans) cout << x + 1 << " ";
  cout << "\n";
}
