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 save[1010][5];
int main(void) {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 5; j++) scanf("%d", &save[i][j]);
  if (n > 11) {
    printf("0\n");
    return 0;
  }
  if (n <= 2) {
    printf("%d\n", n);
    printf("%s\n", n == 2 ? "1\n2\n" : "1\n");
    return 0;
  }
  int goodidx = -1;
  for (int i = 0; i < n; i++) {
    int flag = 1;
    for (int j = 0; j < n; j++) {
      if (j == i) continue;
      for (int k = j + 1; k < n; k++) {
        if (i == k) continue;
        int pp = 0;
        for (int x = 0; x < 5; x++)
          pp += (save[j][x] - save[i][x]) * (save[k][x] - save[i][x]);
        if (pp > 0) flag = 0;
      }
      if (flag == 0) break;
    }
    if (flag) goodidx = i;
    if (goodidx != -1) {
      printf("1\n%d\n", goodidx + 1);
      return 0;
    }
  }
  printf("0\n");
  return 0;
}
