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;
vector<int> a, b, c, d, e;
unsigned long long pr(vector<int> x, vector<int> y) {
  return x[0] * y[0] + x[1] * y[1] + x[2] * y[2] + x[3] * y[3] + x[4] * y[4];
}
vector<int> vec(int x, int y) {
  return {a[y] - a[x], b[y] - b[x], c[y] - c[x], d[y] - d[x], e[y] - e[x]};
}
int main() {
  int n;
  cin >> n;
  map<int, bool> bad;
  a = b = c = d = e = vector<int>(n);
  bool is_ok = false;
  double pi = acos(-1);
  for (int i = 0; i < n; ++i) cin >> a[i] >> b[i] >> c[i] >> d[i] >> e[i];
  for (int i = 0; i < n; ++i) {
    is_ok = false;
    for (int j = 0; j < n && !is_ok; ++j) {
      if (j == i) continue;
      for (int k = 0; k < n && !is_ok; ++k) {
        if (k == j || k == i) continue;
        if (acos(pr(vec(i, j), vec(i, k)) / (sqrt(pr(vec(i, j), vec(i, j))) *
                                             sqrt(pr(vec(i, k), vec(i, k))))) <
            acos(-1) / 2) {
          bad[i] = true;
          is_ok = true;
        }
      }
    }
  }
  cout << n - bad.size() << endl;
  for (int i = 0; i < n; ++i) {
    if (!bad[i]) cout << i + 1 << endl;
  }
  return 0;
}
