#include <bits/stdc++.h>
using namespace std;
int n, a[1001][10], cnt, ans[1001];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d%d%d%d", &a[i][0], &a[i][1], &a[i][2], &a[i][3], &a[i][4]);
  }
  if (n > 11) {
    printf("0");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    bool flg = 0;
    for (int j = 1; j <= n; j++)
      if (i != j)
        for (int k = j + 1; k <= n; k++)
          if (k != i) {
            int mx = 0;
            double mmx = 0, mmy = 0;
            for (int p = 0; p < 5; p++) {
              mx += (a[i][p] - a[j][p]) * (a[i][p] - a[k][p]);
              mmx += (a[i][p] - a[j][p]) * (a[i][p] - a[j][p]);
              mmy += (a[i][p] - a[k][p]) * (a[i][p] - a[k][p]);
            }
            mmx = sqrt(mmx);
            mmy = sqrt(mmy);
            if (acos((double)mx / (mmx * mmy)) * 2 <= acos(-1) - 1e-8) flg = 1;
          }
    if (!flg) ans[++cnt] = i;
  }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) printf("%d\n", ans[i]);
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
int n, a[1001][10], cnt, ans[1001];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d%d%d%d", &a[i][0], &a[i][1], &a[i][2], &a[i][3], &a[i][4]);
  }
  if (n > 11) {
    printf("0");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    bool flg = 0;
    for (int j = 1; j <= n; j++)
      if (i != j)
        for (int k = j + 1; k <= n; k++)
          if (k != i) {
            int mx = 0;
            double mmx = 0, mmy = 0;
            for (int p = 0; p < 5; p++) {
              mx += (a[i][p] - a[j][p]) * (a[i][p] - a[k][p]);
              mmx += (a[i][p] - a[j][p]) * (a[i][p] - a[j][p]);
              mmy += (a[i][p] - a[k][p]) * (a[i][p] - a[k][p]);
            }
            mmx = sqrt(mmx);
            mmy = sqrt(mmy);
            if (acos((double)mx / (mmx * mmy)) * 2 <= acos(-1) - 1e-8) flg = 1;
          }
    if (!flg) ans[++cnt] = i;
  }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) printf("%d\n", ans[i]);
}

```