#include <bits/stdc++.h>
using namespace std;
int a[1005][5];
int mu(int a[], int b[]) {
  int res = 0;
  for (int i = 0; i < 5; ++i) {
    res += a[i] * b[i];
  }
  return res;
}
bool check(int d1, int d2, int k1, int k2) {
  int d[5], k[5];
  for (int i = 0; i < 5; ++i) {
    d[i] = a[d2][i] - a[d1][i];
    k[i] = a[k2][i] - a[k1][i];
  }
  return mu(d, k) > 0;
}
int main(int argc, const char* argv[]) {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < 5; ++j) {
      scanf("%d", a[i] + j);
    }
  }
  if (n > 500)
    puts("0");
  else {
    vector<int> ans;
    for (int i = 1; i <= n; ++i) {
      bool f = false;
      for (int j = 1; !f && j <= n; ++j) {
        if (j == i) continue;
        for (int k = 1; !f && k <= n; ++k) {
          if (k == i || k == j) continue;
          if (check(i, j, i, k)) {
            f = true;
          }
        }
      }
      if (!f) ans.push_back(i);
    }
    cout << ans.size() << endl;
    for (int x : ans) {
      cout << x << ' ';
    }
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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;
int a[1005][5];
int mu(int a[], int b[]) {
  int res = 0;
  for (int i = 0; i < 5; ++i) {
    res += a[i] * b[i];
  }
  return res;
}
bool check(int d1, int d2, int k1, int k2) {
  int d[5], k[5];
  for (int i = 0; i < 5; ++i) {
    d[i] = a[d2][i] - a[d1][i];
    k[i] = a[k2][i] - a[k1][i];
  }
  return mu(d, k) > 0;
}
int main(int argc, const char* argv[]) {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < 5; ++j) {
      scanf("%d", a[i] + j);
    }
  }
  if (n > 500)
    puts("0");
  else {
    vector<int> ans;
    for (int i = 1; i <= n; ++i) {
      bool f = false;
      for (int j = 1; !f && j <= n; ++j) {
        if (j == i) continue;
        for (int k = 1; !f && k <= n; ++k) {
          if (k == i || k == j) continue;
          if (check(i, j, i, k)) {
            f = true;
          }
        }
      }
      if (!f) ans.push_back(i);
    }
    cout << ans.size() << endl;
    for (int x : ans) {
      cout << x << ' ';
    }
  }
  return 0;
}

```