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 main() {
  int point[1010][5], dis[2][5], ans[20], n;
  int i, j, k, p, sum, flag, key;
  scanf("%d", &n);
  for (i = 0; i < n; i++)
    for (j = 0; j < 5; j++) scanf("%d", &point[i][j]);
  if (n > 11)
    printf("0\n");
  else {
    key = 0;
    for (i = 0; i < n; i++) {
      flag = 1;
      for (j = 0; j < n; j++) {
        for (k = 0; k < n; k++) {
          sum = 0;
          if (i == j || i == k || j == k)
            continue;
          else {
            for (p = 0; p < 5; p++) {
              dis[0][p] = point[j][p] - point[i][p];
              dis[1][p] = point[k][p] - point[i][p];
            }
            for (p = 0; p < 5; p++) sum += dis[0][p] * dis[1][p];
            if (sum > 0) {
              flag = 0;
              break;
            }
          }
        }
        if (!flag) break;
      }
      if (flag) {
        ans[key] = i + 1;
        key++;
      }
    }
    printf("%d\n", key);
    for (i = 0; i < key; i++) printf("%d\n", ans[i]);
  }
  return 0;
}
