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 ans[10010];
bool vis[10010];
struct node {
  int x1, x2, x3, x4, x5;
} a[10010];
int main() {
  int n, cnt = 0;
  int f = 0;
  cin >> n;
  for (int i = 1; i <= n; i++)
    cin >> a[i].x1 >> a[i].x2 >> a[i].x3 >> a[i].x4 >> a[i].x5;
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    f = 0;
    for (int j = 1; j <= n; j++) {
      if (j == i) continue;
      if (f == 1) break;
      for (int k = 1; k <= n; k++) {
        if (k == i || k == j) continue;
        int dx1 = a[j].x1 - a[i].x1;
        int dx2 = a[j].x2 - a[i].x2;
        int dx3 = a[j].x3 - a[i].x3;
        int dx4 = a[j].x4 - a[i].x4;
        int dx5 = a[j].x5 - a[i].x5;
        int cx1 = a[k].x1 - a[i].x1;
        int cx2 = a[k].x2 - a[i].x2;
        int cx3 = a[k].x3 - a[i].x3;
        int cx4 = a[k].x4 - a[i].x4;
        int cx5 = a[k].x5 - a[i].x5;
        int t = dx1 * cx1 + dx2 * cx2 + dx3 * cx3 + dx4 * cx4 + dx5 * cx5;
        if (t > 0) {
          f = 1;
          break;
        } else {
          vis[k] = 1;
          vis[j] = 1;
        }
        if (f == 1) break;
      }
    }
    if (f == 0) ans[++cnt] = i;
  }
  cout << cnt << endl;
  for (int i = 1; i < cnt; i++) cout << ans[i] << ' ';
  if (cnt) cout << ans[cnt] << endl;
}
