#include <bits/stdc++.h>
using namespace std;
const long long inf = 1LL << 61;
const long long mod = 1000000007;
const int maxn = 1010;
double a[maxn], b[maxn], c[maxn], d[maxn], e[maxn];
int main() {
  std::ios::sync_with_stdio(false);
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%lf%lf%lf%lf%lf", &a[i], &b[i], &c[i], &d[i], &e[i]);
  }
  if (n >= 12) {
    puts("0");
    return 0;
  }
  int k = 0;
  vector<int> ans;
  for (int i = 0; i < n; i++) {
    bool flag = false;
    for (int j = 0; j < n; j++) {
      if (j == i) continue;
      for (int s = j + 1; s < n; s++) {
        if (s == i) continue;
        double v =
            (a[j] - a[i]) * (a[s] - a[i]) + (b[j] - b[i]) * (b[s] - b[i]) +
            (c[j] - c[i]) * (c[s] - c[i]) + (d[j] - d[i]) * (d[s] - d[i]) +
            (e[j] - e[i]) * (e[s] - e[i]);
        if (v > 0) {
          flag = true;
          break;
        }
      }
      if (flag) break;
    }
    if (!flag) {
      k++;
      ans.push_back(i + 1);
    }
  }
  printf("%d\n", k);
  for (int i = 0; i < k; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution to 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;
const long long inf = 1LL << 61;
const long long mod = 1000000007;
const int maxn = 1010;
double a[maxn], b[maxn], c[maxn], d[maxn], e[maxn];
int main() {
  std::ios::sync_with_stdio(false);
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%lf%lf%lf%lf%lf", &a[i], &b[i], &c[i], &d[i], &e[i]);
  }
  if (n >= 12) {
    puts("0");
    return 0;
  }
  int k = 0;
  vector<int> ans;
  for (int i = 0; i < n; i++) {
    bool flag = false;
    for (int j = 0; j < n; j++) {
      if (j == i) continue;
      for (int s = j + 1; s < n; s++) {
        if (s == i) continue;
        double v =
            (a[j] - a[i]) * (a[s] - a[i]) + (b[j] - b[i]) * (b[s] - b[i]) +
            (c[j] - c[i]) * (c[s] - c[i]) + (d[j] - d[i]) * (d[s] - d[i]) +
            (e[j] - e[i]) * (e[s] - e[i]);
        if (v > 0) {
          flag = true;
          break;
        }
      }
      if (flag) break;
    }
    if (!flag) {
      k++;
      ans.push_back(i + 1);
    }
  }
  printf("%d\n", k);
  for (int i = 0; i < k; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

```