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 = 1000 + 7;
int a[MAXN], b[MAXN], c[MAXN], d[MAXN], e[MAXN];
int isGood[MAXN];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i)
    scanf("%d%d%d%d%d", &a[i], &b[i], &c[i], &d[i], &e[i]);
  memset(isGood, -1, sizeof(isGood));
  int cnt = n;
  if (n >= 243) {
    printf("0\n");
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      for (int k = 1; k <= n; ++k) {
        if ((j == i) || (j == k) || (i == k)) continue;
        long long a1 = a[j] - a[i], b1 = b[j] - b[i], c1 = c[j] - c[i],
                  d1 = d[j] - d[i];
        long long e1 = e[j] - e[i];
        long long a2 = a[k] - a[i], b2 = b[k] - b[i], c2 = c[k] - c[i],
                  d2 = d[k] - d[i];
        long long e2 = e[k] - e[i];
        long long res = a1 * a2 + b1 * b2 + c1 * c2 + d1 * d2 + e1 * e2;
        if (res > 0) {
          if (isGood[i]) isGood[i] = 0, cnt--;
        }
      }
    }
  }
  printf("%d\n", cnt);
  for (int i = 1; i <= n; ++i)
    if (isGood[i]) printf("%d\n", i);
  return 0;
}
