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[5], id;
} p[1005];
inline Point operator-(Point a, Point b) {
  Point tmp;
  for (int i = 0; i < 5; i++) tmp.a[i] = a.a[i] - b.a[i];
  return tmp;
}
inline int cross(Point a, Point b) {
  int tmp = 0;
  for (int i = 0; i < 5; i++) tmp += a.a[i] * b.a[i];
  return tmp;
}
queue<Point> Q;
int n;
vector<int> ans;
inline bool judge(Point a, Point b, Point c) {
  return cross(b - a, c - a) <= 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 5; j++) scanf("%d", &p[i].a[j]);
    p[i].id = i;
    Q.push(p[i]);
  }
  while (Q.size() >= 3) {
    Point a = Q.front();
    Q.pop();
    Point b = Q.front();
    Q.pop();
    Point c = Q.front();
    Q.pop();
    if (judge(a, b, c)) Q.push(a);
    if (judge(b, a, c)) Q.push(b);
    if (judge(c, a, b)) Q.push(c);
  }
  while (!Q.empty()) {
    Point a = Q.front();
    Q.pop();
    bool flag = true;
    for (int i = 1; i <= n; i++)
      if (i != a.id)
        for (int j = 1; j <= n; j++)
          if (i != j && j != a.id)
            if (!judge(a, p[i], p[j])) flag = false;
    if (flag) ans.push_back(a.id);
  }
  printf("%d\n", (int)ans.size());
  for (int i = 0; i < ans.size(); i++) printf("%d\n", ans[i]);
  return 0;
}
