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;
long long a[105][5];
int n, Ans, b[105];
bool check(int i) {
  for (int j = 1; j <= n; j++)
    for (int k = j + 1; k <= n; k++)
      if (i != j && i != k) {
        long long tmp = 0;
        for (int l = 1; l <= 5; l++)
          tmp += (a[j][l] - a[i][l]) * (a[k][l] - a[i][l]);
        if (tmp > 0) return 0;
      }
  return 1;
}
int main() {
  scanf("%d", &n);
  if (n > 66) {
    puts("0");
  } else {
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= 5; j++) scanf("%I64d", &a[i][j]);
    for (int i = 1; i <= n; i++) {
      Ans += b[i] = check(i);
    }
    printf("%d\n", Ans);
    for (int i = 1; i <= n; i++)
      if (b[i]) {
        printf("%d\n", i);
      }
  }
}
