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 Nmax = 1e3 + 7;
const double PI = acos(-1.0);
int n;
struct Point {
  int a[6];
} p[Nmax];
int is[Nmax];
Point getv(Point a, Point b) {
  Point tmp;
  for (int i = 1; i <= 5; i++) tmp.a[i] = b.a[i] - a.a[i];
  return tmp;
}
double len(Point x, Point y) {
  double ans = 0.0;
  for (int i = 1; i <= 5; i++) ans += x.a[i] * y.a[i];
  return ans;
}
double get(Point x, Point y) {
  double a = len(x, y), b = sqrt(len(x, x)) * sqrt(len(y, y));
  return acos(a / b);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= 5; j++) scanf("%d", &p[i].a[j]);
  if (n > 243) {
    printf("0\n");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      for (int k = j + 1; k <= n; k++) {
        double now = get(getv(p[i], p[j]), getv(p[i], p[k]));
        if (now < PI / 2.0) is[i] = 1;
      }
    }
  }
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (!is[i]) cnt++;
  }
  printf("%d\n", cnt);
  for (int i = 1; i <= n; i++) {
    if (!is[i]) printf("%d\n", i);
  }
  return 0;
}
