#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 3;
int n, Pnt[N][6], res[N], tot;
int main() {
  scanf("%d", &n);
  for (register int i = (1); i <= (n); ++i)
    for (register int j = (1); j <= (5); ++j) scanf("%d", Pnt[i] + j);
  if (n > 34) return puts("0"), 0;
  for (register int i = (1); i <= (n); ++i) {
    for (register int j = (1); j <= (n); ++j)
      if (j != i)
        for (register int k = (1); k <= (n); ++k)
          if (j != k && i != k) {
            int res = 0;
            for (register int p = (1); p <= (5); ++p)
              res += (Pnt[j][p] - Pnt[i][p]) * (Pnt[k][p] - Pnt[i][p]);
            if (res > 0) goto ED;
          }
    res[++tot] = i;
  ED:;
  }
  printf("%d\n", tot);
  for (register int i = (1); i <= (tot); ++i)
    printf("%d%c", res[i], " \n"[i == tot]);
  return 0;
}

### Prompt

Develop a solution in cpp 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;
const int N = 1e3 + 3;
int n, Pnt[N][6], res[N], tot;
int main() {
  scanf("%d", &n);
  for (register int i = (1); i <= (n); ++i)
    for (register int j = (1); j <= (5); ++j) scanf("%d", Pnt[i] + j);
  if (n > 34) return puts("0"), 0;
  for (register int i = (1); i <= (n); ++i) {
    for (register int j = (1); j <= (n); ++j)
      if (j != i)
        for (register int k = (1); k <= (n); ++k)
          if (j != k && i != k) {
            int res = 0;
            for (register int p = (1); p <= (5); ++p)
              res += (Pnt[j][p] - Pnt[i][p]) * (Pnt[k][p] - Pnt[i][p]);
            if (res > 0) goto ED;
          }
    res[++tot] = i;
  ED:;
  }
  printf("%d\n", tot);
  for (register int i = (1); i <= (tot); ++i)
    printf("%d%c", res[i], " \n"[i == tot]);
  return 0;
}

```