#include <bits/stdc++.h>
using namespace std;
struct Data {
  int a, b, c, d, e;
};
Data a[1100];
int ans[1100], cnt;
int main() {
  int n;
  while (~scanf("%d", &n)) {
    for (int k = 1; k <= n; k++)
      scanf("%d %d %d %d %d", &a[k].a, &a[k].b, &a[k].c, &a[k].d, &a[k].e);
    cnt = 0;
    int bo;
    for (int k = 1; k <= n; k++) {
      bo = 1;
      for (int i = 1; i <= n; i++) {
        if (bo == 0) break;
        if (i == k) continue;
        for (int j = i + 1; j <= n; j++) {
          if (j == k) continue;
          int save = (a[j].a - a[k].a) * (a[i].a - a[k].a) +
                     (a[j].b - a[k].b) * (a[i].b - a[k].b) +
                     (a[j].c - a[k].c) * (a[i].c - a[k].c) +
                     (a[j].d - a[k].d) * (a[i].d - a[k].d) +
                     (a[j].e - a[k].e) * (a[i].e - a[k].e);
          if (save > 0) {
            bo = 0;
            break;
          }
        }
      }
      if (bo) {
        cnt++;
        ans[cnt] = k;
      }
    }
    printf("%d\n", cnt);
    for (int k = 1; k <= cnt; k++) printf("%d\n", ans[k]);
  }
  return 0;
}

### Prompt

Please formulate a cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
struct Data {
  int a, b, c, d, e;
};
Data a[1100];
int ans[1100], cnt;
int main() {
  int n;
  while (~scanf("%d", &n)) {
    for (int k = 1; k <= n; k++)
      scanf("%d %d %d %d %d", &a[k].a, &a[k].b, &a[k].c, &a[k].d, &a[k].e);
    cnt = 0;
    int bo;
    for (int k = 1; k <= n; k++) {
      bo = 1;
      for (int i = 1; i <= n; i++) {
        if (bo == 0) break;
        if (i == k) continue;
        for (int j = i + 1; j <= n; j++) {
          if (j == k) continue;
          int save = (a[j].a - a[k].a) * (a[i].a - a[k].a) +
                     (a[j].b - a[k].b) * (a[i].b - a[k].b) +
                     (a[j].c - a[k].c) * (a[i].c - a[k].c) +
                     (a[j].d - a[k].d) * (a[i].d - a[k].d) +
                     (a[j].e - a[k].e) * (a[i].e - a[k].e);
          if (save > 0) {
            bo = 0;
            break;
          }
        }
      }
      if (bo) {
        cnt++;
        ans[cnt] = k;
      }
    }
    printf("%d\n", cnt);
    for (int k = 1; k <= cnt; k++) printf("%d\n", ans[k]);
  }
  return 0;
}

```