#include <bits/stdc++.h>
using namespace std;
long long a[1006][6];
int n, vis[1003];
set<long long> s;
bool check(int x, int y, int z) {
  long long pos = 0;
  for (int i = 0; i < 5; i++) pos += (a[y][i] - a[x][i]) * (a[z][i] - a[x][i]);
  if (pos > 0) return false;
  return true;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    s.insert(i + 1);
    for (int j = 0; j < 5; j++) scanf("%lld", &a[i][j]);
  }
  int l = n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      for (int k = j + 1; k < n; k++) {
        if (i != j && i != k) {
          if (!check(i, j, k)) {
            l--;
            s.erase(i + 1);
            goto eg;
          }
        }
      }
    }
  eg:;
  }
  printf("%d\n", l);
  if (s.size()) {
    for (set<long long>::iterator it = s.begin(); it != s.end(); it++) {
      printf("%lld\n", *it);
    }
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
long long a[1006][6];
int n, vis[1003];
set<long long> s;
bool check(int x, int y, int z) {
  long long pos = 0;
  for (int i = 0; i < 5; i++) pos += (a[y][i] - a[x][i]) * (a[z][i] - a[x][i]);
  if (pos > 0) return false;
  return true;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    s.insert(i + 1);
    for (int j = 0; j < 5; j++) scanf("%lld", &a[i][j]);
  }
  int l = n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      for (int k = j + 1; k < n; k++) {
        if (i != j && i != k) {
          if (!check(i, j, k)) {
            l--;
            s.erase(i + 1);
            goto eg;
          }
        }
      }
    }
  eg:;
  }
  printf("%d\n", l);
  if (s.size()) {
    for (set<long long>::iterator it = s.begin(); it != s.end(); it++) {
      printf("%lld\n", *it);
    }
  }
  return 0;
}

```