#include <bits/stdc++.h>
using namespace std;
int n;
vector<int> vet[130];
vector<int> operator-(vector<int> a, vector<int> b) {
  vector<int> c(5);
  for (int(i) = (0); (i) < (5); (i)++) c[i] = a[i] - b[i];
  return c;
}
long long mul(vector<int> a, vector<int> b) {
  long long ret = 0;
  for (int(i) = (0); (i) < (5); (i)++) ret += a[i] * b[i];
  return ret;
}
int main() {
  scanf("%d", &n);
  if (n > 70) {
    puts("0");
    return 0;
  }
  for (int(i) = (0); (i) < (n); (i)++) {
    vet[i].resize(5);
    for (int(j) = (0); (j) < (5); (j)++) scanf("%d", &vet[i][j]);
  }
  vector<int> ans;
  for (int(i) = (0); (i) < (n); (i)++) {
    int ok = 1;
    for (int(j) = (0); (j) < (n); (j)++)
      if (j != i)
        for (int(k) = (0); (k) < (n); (k)++)
          if (k != i && k != j) {
            long long v = mul(vet[j] - vet[i], vet[k] - vet[i]);
            if (v > 0) ok = 0;
          }
    if (ok) ans.push_back(i + 1);
  }
  printf("%d\n", ((int)(ans).size()));
  for (int(i) = (0); (i) < (((int)(ans).size())); (i)++)
    printf("%d%c", ans[i], i == ((int)(ans).size()) - 1 ? '\n' : ' ');
}

### 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;
int n;
vector<int> vet[130];
vector<int> operator-(vector<int> a, vector<int> b) {
  vector<int> c(5);
  for (int(i) = (0); (i) < (5); (i)++) c[i] = a[i] - b[i];
  return c;
}
long long mul(vector<int> a, vector<int> b) {
  long long ret = 0;
  for (int(i) = (0); (i) < (5); (i)++) ret += a[i] * b[i];
  return ret;
}
int main() {
  scanf("%d", &n);
  if (n > 70) {
    puts("0");
    return 0;
  }
  for (int(i) = (0); (i) < (n); (i)++) {
    vet[i].resize(5);
    for (int(j) = (0); (j) < (5); (j)++) scanf("%d", &vet[i][j]);
  }
  vector<int> ans;
  for (int(i) = (0); (i) < (n); (i)++) {
    int ok = 1;
    for (int(j) = (0); (j) < (n); (j)++)
      if (j != i)
        for (int(k) = (0); (k) < (n); (k)++)
          if (k != i && k != j) {
            long long v = mul(vet[j] - vet[i], vet[k] - vet[i]);
            if (v > 0) ok = 0;
          }
    if (ok) ans.push_back(i + 1);
  }
  printf("%d\n", ((int)(ans).size()));
  for (int(i) = (0); (i) < (((int)(ans).size())); (i)++)
    printf("%d%c", ans[i], i == ((int)(ans).size()) - 1 ? '\n' : ' ');
}

```