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 node {
  int a, b, c, d, e;
} a[1244];
int b[1234];
int n;
int main() {
  cin >> n;
  memset(b, 0, sizeof(b));
  for (int i = 1; i <= n; i++) {
    cin >> a[i].a >> a[i].b >> a[i].c >> a[i].d >> a[i].e;
  }
  int ans = 0;
  if (n > 11)
    printf("0\n");
  else {
    for (int i = 1; i <= n; i++) {
      int flag = 1;
      for (int j = 1; j <= n; j++) {
        int aa = a[i].a - a[j].a;
        int bb = a[i].b - a[j].b;
        int cc = a[i].c - a[j].c;
        int dd = a[i].d - a[j].d;
        int ee = a[i].e - a[j].e;
        for (int k = 1; k <= n; k++) {
          if (i != j && j != k && k != i) {
            int aaa = a[i].a - a[k].a;
            int bbb = a[i].b - a[k].b;
            int ccc = a[i].c - a[k].c;
            int ddd = a[i].d - a[k].d;
            int eee = a[i].e - a[k].e;
            int ans1 = aa * aaa + bb * bbb + cc * ccc + dd * ddd + ee * eee;
            if (ans1 <= 0) {
              continue;
            } else {
              flag = 0;
              break;
            }
          } else
            continue;
        }
        if (flag == 0) break;
      }
      if (flag) {
        ans++;
        b[i]++;
      }
    }
    printf("%d\n", ans);
    for (int i = 1; i <= n; i++) {
      if (b[i]) printf("%d\n", i);
    }
  }
  return 0;
}
