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 double pi = acos(-1.0);
const double E = exp(1.0);
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int maxn = 9 + 2 * 1e5;
const int maxm = 9 + 1e5;
int n;
int no[1003][10];
int cnt;
bool vis[1003];
int main() {
  scanf("%d", &n);
  for (int(i) = (0); (i) < (n); (i)++) {
    for (int(j) = (0); (j) < (5); (j)++) {
      scanf("%d", &no[i][j]);
    }
  }
  cnt = n;
  memset(vis, 1, sizeof(vis));
  ;
  for (int(i) = (0); (i) < (n); (i)++) {
    for (int(j) = (0); (j) < (n); (j)++) {
      if (j == i) continue;
      for (int(k) = (0); (k) < (n); (k)++) {
        if (k == i) continue;
        if (k == j) continue;
        if ((no[j][0] - no[i][0]) * (no[k][0] - no[i][0]) +
                (no[j][1] - no[i][1]) * (no[k][1] - no[i][1]) +
                (no[j][2] - no[i][2]) * (no[k][2] - no[i][2]) +
                (no[j][3] - no[i][3]) * (no[k][3] - no[i][3]) +
                (no[j][4] - no[i][4]) * (no[k][4] - no[i][4]) >
            0) {
          vis[i] = 0;
          cnt--;
          break;
        }
      }
      if (vis[i] == 0) break;
    }
  }
  printf("%d\n", cnt);
  bool tag = 1;
  for (int(i) = (0); (i) < (n); (i)++) {
    if (vis[i]) {
      if (tag) {
        tag = 0;
        printf("%d", i + 1);
      } else
        printf(" %d", i + 1);
    }
  }
  return 0;
}
