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 Point {
  int a, b, c, d, e;
  Point operator-(const Point& o) const {
    return {a - o.a, b - o.b, c - o.c, d - o.d, e - o.e};
  }
  int operator*(const Point& o) const {
    return a * o.a + b * o.b + c * o.c + d * o.d + e * o.e;
  }
  friend istream& operator>>(istream& s, Point& p) {
    return s >> p.a >> p.b >> p.c >> p.d >> p.e;
  }
};
Point points[1000];
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> good;
  good.reserve(n);
  for (int i = 0; i < n; ++i) cin >> points[i];
  for (int i = 0; i < n; ++i) {
    bool g = true;
    for (int j = 0; j < n; ++j)
      if (j != i)
        for (int k = j + 1; k < n; ++k)
          if (k != i) {
            Point a = points[j] - points[i];
            Point b = points[k] - points[i];
            if (a * b > 0) {
              g = false;
              break;
            }
          }
    if (g) good.push_back(i);
  }
  cout << good.size() << '\n';
  for (int x : good) cout << x + 1 << ' ';
  if (good.size()) cout << '\n';
  return 0;
}
