#include <bits/stdc++.h>
using namespace std;
const int MAX = 1010;
struct point {
  int x, y, z, u, w;
} in[MAX];
bool bad[MAX];
int a, b, c;
int scalar(int i, int j, int k) {
  return ((in[i].x - in[k].x) * (in[j].x - in[k].x) +
          (in[i].y - in[k].y) * (in[j].y - in[k].y) +
          (in[i].z - in[k].z) * (in[j].z - in[k].z) +
          (in[i].u - in[k].u) * (in[j].u - in[k].u) +
          (in[i].w - in[k].w) * (in[j].w - in[k].w));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  vector<int> ans;
  stack<int> s;
  int n;
  cin >> n;
  if (n == 0) {
    cout << 0 << endl;
    return 0;
  }
  for (int i = 0; i < n; ++i) {
    cin >> in[i].x >> in[i].y >> in[i].z >> in[i].u >> in[i].w;
    s.push(i);
  }
  while (!s.empty()) {
    if (s.size() > 2) {
      a = s.top();
      s.pop();
      b = s.top();
      s.pop();
      c = s.top();
      s.pop();
      if (scalar(a, b, c) <= 0) {
        bad[a] = true;
        bad[b] = true;
        s.push(c);
      } else if (scalar(c, a, b) <= 0) {
        bad[c] = true;
        bad[a] = true;
        s.push(b);
      } else if (scalar(b, c, a) <= 0) {
        bad[b] = true;
        bad[c] = true;
        s.push(a);
      } else {
        bad[a] = true;
        bad[b] = true;
        bad[c] = true;
      }
    } else {
      a = s.top();
      s.pop();
      bool flag = true;
      for (int i = 0; i < n; ++i) {
        if (i == a) continue;
        for (int j = 0; j < n; ++j) {
          if (i == j || j == a) continue;
          if (scalar(i, j, a) > 0) flag = false;
        }
      }
      if (flag)
        ans.push_back(a);
      else
        bad[a] = true;
    }
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); ++i) cout << (ans[i] + 1) << endl;
  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;
const int MAX = 1010;
struct point {
  int x, y, z, u, w;
} in[MAX];
bool bad[MAX];
int a, b, c;
int scalar(int i, int j, int k) {
  return ((in[i].x - in[k].x) * (in[j].x - in[k].x) +
          (in[i].y - in[k].y) * (in[j].y - in[k].y) +
          (in[i].z - in[k].z) * (in[j].z - in[k].z) +
          (in[i].u - in[k].u) * (in[j].u - in[k].u) +
          (in[i].w - in[k].w) * (in[j].w - in[k].w));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  vector<int> ans;
  stack<int> s;
  int n;
  cin >> n;
  if (n == 0) {
    cout << 0 << endl;
    return 0;
  }
  for (int i = 0; i < n; ++i) {
    cin >> in[i].x >> in[i].y >> in[i].z >> in[i].u >> in[i].w;
    s.push(i);
  }
  while (!s.empty()) {
    if (s.size() > 2) {
      a = s.top();
      s.pop();
      b = s.top();
      s.pop();
      c = s.top();
      s.pop();
      if (scalar(a, b, c) <= 0) {
        bad[a] = true;
        bad[b] = true;
        s.push(c);
      } else if (scalar(c, a, b) <= 0) {
        bad[c] = true;
        bad[a] = true;
        s.push(b);
      } else if (scalar(b, c, a) <= 0) {
        bad[b] = true;
        bad[c] = true;
        s.push(a);
      } else {
        bad[a] = true;
        bad[b] = true;
        bad[c] = true;
      }
    } else {
      a = s.top();
      s.pop();
      bool flag = true;
      for (int i = 0; i < n; ++i) {
        if (i == a) continue;
        for (int j = 0; j < n; ++j) {
          if (i == j || j == a) continue;
          if (scalar(i, j, a) > 0) flag = false;
        }
      }
      if (flag)
        ans.push_back(a);
      else
        bad[a] = true;
    }
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); ++i) cout << (ans[i] + 1) << endl;
  return 0;
}

```