#include <bits/stdc++.h>
using namespace std;
void doRoutine() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
struct pt {
  vector<int64_t> a;
  pt() {}
  pt(vector<int64_t> b) { a = b; }
  int64_t get(int64_t i) { return a[i]; }
};
int64_t scc(pt a, pt b, pt c) {
  int64_t res = 0;
  for (int64_t i = 0; i < 5; ++i) {
    res += (b.get(i) - a.get(i)) * (c.get(i) - a.get(i));
  }
  return res;
}
signed main() {
  int64_t n;
  cin >> n;
  if (n >= 20) {
    cout << 0;
    return 0;
  }
  vector<pt> p(n);
  for (int64_t i = 0; i < n; ++i) {
    vector<int64_t> cur(5);
    for (int64_t j = 0; j < 5; ++j) {
      cin >> cur[j];
    }
    p[i] = pt(cur);
  }
  vector<int64_t> ans;
  for (int64_t i = 0; i < n; ++i) {
    bool flag = true;
    for (int64_t j = 0; j < n; ++j) {
      for (int64_t k = 0; k < n; ++k) {
        if (i != j && j != k && i != k) {
          int64_t tmp = scc(p[i], p[j], p[k]);
          if (tmp > 0) {
            flag = false;
          }
        }
      }
    }
    if (flag) {
      ans.push_back(i + 1);
    }
  }
  cout << ans.size() << endl;
  for (auto x : ans) {
    cout << x << 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;
void doRoutine() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
struct pt {
  vector<int64_t> a;
  pt() {}
  pt(vector<int64_t> b) { a = b; }
  int64_t get(int64_t i) { return a[i]; }
};
int64_t scc(pt a, pt b, pt c) {
  int64_t res = 0;
  for (int64_t i = 0; i < 5; ++i) {
    res += (b.get(i) - a.get(i)) * (c.get(i) - a.get(i));
  }
  return res;
}
signed main() {
  int64_t n;
  cin >> n;
  if (n >= 20) {
    cout << 0;
    return 0;
  }
  vector<pt> p(n);
  for (int64_t i = 0; i < n; ++i) {
    vector<int64_t> cur(5);
    for (int64_t j = 0; j < 5; ++j) {
      cin >> cur[j];
    }
    p[i] = pt(cur);
  }
  vector<int64_t> ans;
  for (int64_t i = 0; i < n; ++i) {
    bool flag = true;
    for (int64_t j = 0; j < n; ++j) {
      for (int64_t k = 0; k < n; ++k) {
        if (i != j && j != k && i != k) {
          int64_t tmp = scc(p[i], p[j], p[k]);
          if (tmp > 0) {
            flag = false;
          }
        }
      }
    }
    if (flag) {
      ans.push_back(i + 1);
    }
  }
  cout << ans.size() << endl;
  for (auto x : ans) {
    cout << x << endl;
  }
  return 0;
}

```