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 debugger {
  template <typename T>
  debugger& operator,(const T& v) {
    cerr << v << " ";
    return *this;
  }
} dbg;
struct PT {
  int a, b, c, d, e, id;
};
bool check(PT p1, PT p2, PT p3) {
  int x = (p2.a - p1.a) * (p3.a - p1.a) + (p2.b - p1.b) * (p3.b - p1.b) +
          (p2.c - p1.c) * (p3.c - p1.c) + (p2.d - p1.d) * (p3.d - p1.d) +
          (p2.e - p1.e) * (p3.e - p1.e);
  return x > 0;
}
int main() {
  int n;
  scanf("%d", &n);
  PT p[n + 5];
  for (int i = 1; i <= n; i++) {
    int a, b, c, d, e;
    scanf("%d %d %d %d", &a, &b, &c, &d);
    scanf("%d", &e);
    p[i].a = a;
    p[i].b = b;
    p[i].c = c;
    p[i].d = d;
    p[i].e = e;
    p[i].id = i;
  }
  int taken[n + 5];
  memset(taken, 0, sizeof(taken));
  for (int i = 1; i <= n; i++) {
    if (!taken[i]) {
      for (int j = 1; j <= n; j++) {
        for (int k = 1; k <= n; k++) {
          if (i != j && i != k && j != k) {
            if (check(p[i], p[j], p[k])) taken[i] = 1;
            if (check(p[j], p[i], p[k])) taken[j] = 1;
            if (check(p[k], p[j], p[i])) taken[k] = 1;
          }
        }
      }
    }
  }
  vector<int> ans;
  for (int i = 1; i <= n; i++)
    if (!taken[i]) ans.push_back(i);
  printf("%d\n", (int)ans.size());
  for (int i = 0; i < (int)ans.size(); i++) printf("%d ", ans[i]);
  if ((int)ans.size()) printf("\n");
  return 0;
}
