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 N = 2e5 + 5;
struct Point {
  double a, b, c, d, e;
  Point(double va, double vb, double vc, double vd, double ve) {
    a = va, b = vb, c = vc, d = vd, e = ve;
  }
  Point() {}
} p[N];
Point operator-(const Point& v1, const Point& v2) {
  return Point(v2.a - v1.a, v2.b - v1.b, v2.c - v1.c, v2.d - v1.d, v2.e - v1.e);
}
double dot(Point v1, Point v2) {
  return v1.a * v2.a + v1.b * v2.b + v1.c * v2.c + v1.d * v2.d + v1.e * v2.e;
}
int n;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  vector<int> res;
  cin >> n;
  for (int i = 1; i <= n; i++)
    cin >> p[i].a >> p[i].b >> p[i].c >> p[i].d >> p[i].e;
  for (int i = 1; i <= n; i++) {
    bool flag = true;
    for (int j = 1; j <= n && flag; j++) {
      if (i == j) continue;
      for (int k = 1; k <= n; k++) {
        if (i == k || j == k) continue;
        Point v1 = p[j] - p[i], v2 = p[k] - p[i];
        if (dot(v1, v2) > 0) {
          flag = false;
          break;
        }
      }
    }
    if (flag) res.push_back(i);
  }
  printf("%d\n", res.size());
  for (int i = 0; i < res.size(); i++) printf("%d ", res[i]);
  return 0;
}
