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;
int v[1010][10], rez[1010];
int main() {
  int n, nr = 0, l = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int p = 0;
    for (int j = 1; j <= 5; j++) scanf("%d", &v[i][j]);
  }
  if (n > 200) {
    printf("0");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    int p = 0;
    for (int j = 1; j <= n; j++)
      if (i != j) {
        for (int k = 1; k <= n; k++)
          if (j != k && i != k) {
            int c = 0;
            for (int m = 1; m <= 5; m++)
              c += (v[j][m] - v[i][m]) * (v[k][m] - v[i][m]);
            if (c > 0) {
              p = 1;
              break;
            }
          }
        if (p == 1) break;
      }
    if (p == 0) rez[++l] = i;
  }
  printf("%d\n", l);
  for (int i = 1; i <= l; i++) printf("%d\n", rez[i]);
  return 0;
}
