#include <bits/stdc++.h>
using namespace std;
const int N = 1e5;
int gi() {
  int w = 0;
  bool q = 1;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') q = 0, c = getchar();
  while (c >= '0' && c <= '9') w = w * 10 + c - '0', c = getchar();
  return q ? w : -w;
}
int a[N], b[N], c[N], d[N], e[N];
inline bool check(int n, int k) {
  for (int i = 1, j; i <= n; i++)
    for (j = i + 1; j <= n; j++)
      if ((a[i] - a[k]) * (a[j] - a[k]) + (b[i] - b[k]) * (b[j] - b[k]) +
              (c[i] - c[k]) * (c[j] - c[k]) + (d[i] - d[k]) * (d[j] - d[k]) +
              (e[i] - e[k]) * (e[j] - e[k]) >
          0)
        return false;
  return true;
}
int main() {
  int n = gi(), i, k;
  for (i = 1; i <= n; i++)
    a[i] = gi(), b[i] = gi(), c[i] = gi(), d[i] = gi(), e[i] = gi();
  vector<int> ans;
  for (k = 1; k <= n; k++)
    if (check(n, k)) ans.push_back(k);
  printf("%d\n", ans.size());
  for (i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 = 1e5;
int gi() {
  int w = 0;
  bool q = 1;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') q = 0, c = getchar();
  while (c >= '0' && c <= '9') w = w * 10 + c - '0', c = getchar();
  return q ? w : -w;
}
int a[N], b[N], c[N], d[N], e[N];
inline bool check(int n, int k) {
  for (int i = 1, j; i <= n; i++)
    for (j = i + 1; j <= n; j++)
      if ((a[i] - a[k]) * (a[j] - a[k]) + (b[i] - b[k]) * (b[j] - b[k]) +
              (c[i] - c[k]) * (c[j] - c[k]) + (d[i] - d[k]) * (d[j] - d[k]) +
              (e[i] - e[k]) * (e[j] - e[k]) >
          0)
        return false;
  return true;
}
int main() {
  int n = gi(), i, k;
  for (i = 1; i <= n; i++)
    a[i] = gi(), b[i] = gi(), c[i] = gi(), d[i] = gi(), e[i] = gi();
  vector<int> ans;
  for (k = 1; k <= n; k++)
    if (check(n, k)) ans.push_back(k);
  printf("%d\n", ans.size());
  for (i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
  return 0;
}

```