#include <bits/stdc++.h>
using namespace std;
int n, good;
long long a[1009][10];
vector<int> ans;
bool solve(int i, int j, int k) {
  long long x[10], y[10];
  for (int d = 1; d <= 5; d++)
    x[d] = a[j][d] - a[i][d], y[d] = a[k][d] - a[i][d];
  long long sum1 = 0, sum2 = 0, sum3 = 0;
  for (int d = 1; d <= 5; d++) {
    sum1 += x[d] * y[d];
    sum2 += x[d] * x[d];
    sum3 += y[d] * y[d];
  }
  if (sum1 < 0) return false;
  sum1 *= sum1;
  return (sum1 <= sum2 * sum3 && sum1 > 0);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++)
    for (int j = 1; j <= 5; j++) cin >> a[i][j];
  for (int i = 0; i < n; i++) {
    good = 0;
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < n; k++)
        if (i != j && i != k && j != k && solve(i, j, k)) {
          good = 1;
          break;
        }
      if (good) break;
    }
    if (!good) ans.push_back(i + 1);
  }
  cout << ans.size() << "\n";
  for (auto i : ans) cout << i << " ";
}

### 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, good;
long long a[1009][10];
vector<int> ans;
bool solve(int i, int j, int k) {
  long long x[10], y[10];
  for (int d = 1; d <= 5; d++)
    x[d] = a[j][d] - a[i][d], y[d] = a[k][d] - a[i][d];
  long long sum1 = 0, sum2 = 0, sum3 = 0;
  for (int d = 1; d <= 5; d++) {
    sum1 += x[d] * y[d];
    sum2 += x[d] * x[d];
    sum3 += y[d] * y[d];
  }
  if (sum1 < 0) return false;
  sum1 *= sum1;
  return (sum1 <= sum2 * sum3 && sum1 > 0);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++)
    for (int j = 1; j <= 5; j++) cin >> a[i][j];
  for (int i = 0; i < n; i++) {
    good = 0;
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < n; k++)
        if (i != j && i != k && j != k && solve(i, j, k)) {
          good = 1;
          break;
        }
      if (good) break;
    }
    if (!good) ans.push_back(i + 1);
  }
  cout << ans.size() << "\n";
  for (auto i : ans) cout << i << " ";
}

```