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 maxn = 1e3 + 10;
const double PI = acos(-1.0);
inline int sign(const double &x) {
  if (x > (1e-10))
    return 1;
  else if (x < -(1e-10))
    return -1;
  return 0;
}
struct Point {
  double a, b, c, d, e;
  Point(double a = 0, double b = 0, double c = 0, double d = 0, double e = 0)
      : a(a), b(b), c(c), d(d), e(e) {}
  Point operator-(Point p) {
    return Point(a - p.a, b - p.b, c - p.c, d - p.d, e - p.e);
  }
} p[maxn];
inline double f1(Point x, Point y) {
  return x.a * y.a + x.b * y.b + x.c * y.c + x.d * y.d + x.e * y.e;
}
inline double f2(Point x) { return sqrt(f1(x, x)); }
inline double getangle(Point x, Point y) {
  return acos(f1(x, y) / f2(x) / f2(y));
}
int main() {
  int n;
  scanf("%d", &n);
  vector<int> ans;
  for (int i = 0; i < n; i++) {
    scanf("%lf%lf%lf%lf%lf", &p[i].a, &p[i].b, &p[i].c, &p[i].d, &p[i].e);
  }
  if (n == 1) {
    printf("1\n");
    printf("1\n");
    return 0;
  }
  if (n == 2) {
    printf("%d\n%d\n%d\n", 2, 1, 2);
    return 0;
  }
  for (int i = 0; i < n; i++) {
    bool flag = false;
    for (int j = 0; j < n && !flag; j++) {
      for (int k = 0; k < n && !flag; k++) {
        if (i == j || i == k || j == k) continue;
        if (sign(getangle(p[j] - p[i], p[k] - p[i]) - PI / 2.0) < 0) {
          flag = true;
          break;
        }
      }
    }
    if (!flag) ans.push_back(i + 1);
  }
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}
