#include <bits/stdc++.h>
using namespace std;
template <class T>
using PIT = pair<int, T>;
template <class T>
using PTI = pair<T, int>;
using PII = pair<int, int>;
using PDI = pair<double, int>;
using LL = long long;
using ULL = unsigned long long;
struct P {
  LL a, b, c, d, e;
};
LL angle(const P &p1, const P &p2) {
  return p1.a * p2.a + p1.b * p2.b + p1.c * p2.c + p1.d * p2.d + p1.e * p2.e;
}
P mainasu(const P &p1, const P &p2) {
  return P{p2.a - p1.a, p2.b - p1.b, p2.c - p1.c, p2.d - p1.d, p2.e - p1.e};
}
int main() {
  std::ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<P> p(n);
  for (int i = 0; i < n; i++) {
    LL a, b, c, d, e;
    cin >> a >> b >> c >> d >> e;
    p[i] = {a, b, c, d, e};
  }
  if (n < 3) {
    cout << n << endl;
    for (int i = 0; i < n; i++) {
      cout << i + 1 << endl;
    }
    return 0;
  }
  vector<int> ans;
  vector<bool> ok(n, true);
  for (int i = 0; i < n; i++) {
    bool flag = true;
    if (!ok[i]) continue;
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < n; k++) {
        if (j == i || k == i || j == k) continue;
        if (angle(mainasu(p[i], p[j]), mainasu(p[i], p[k])) <= 0) {
          ok[j] = false;
          ok[k] = false;
        } else {
          flag = false;
        }
      }
    }
    if (flag) ans.push_back(i + 1);
  }
  cout << ans.size() << endl;
  for (auto anss : ans) {
    cout << anss << endl;
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
template <class T>
using PIT = pair<int, T>;
template <class T>
using PTI = pair<T, int>;
using PII = pair<int, int>;
using PDI = pair<double, int>;
using LL = long long;
using ULL = unsigned long long;
struct P {
  LL a, b, c, d, e;
};
LL angle(const P &p1, const P &p2) {
  return p1.a * p2.a + p1.b * p2.b + p1.c * p2.c + p1.d * p2.d + p1.e * p2.e;
}
P mainasu(const P &p1, const P &p2) {
  return P{p2.a - p1.a, p2.b - p1.b, p2.c - p1.c, p2.d - p1.d, p2.e - p1.e};
}
int main() {
  std::ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<P> p(n);
  for (int i = 0; i < n; i++) {
    LL a, b, c, d, e;
    cin >> a >> b >> c >> d >> e;
    p[i] = {a, b, c, d, e};
  }
  if (n < 3) {
    cout << n << endl;
    for (int i = 0; i < n; i++) {
      cout << i + 1 << endl;
    }
    return 0;
  }
  vector<int> ans;
  vector<bool> ok(n, true);
  for (int i = 0; i < n; i++) {
    bool flag = true;
    if (!ok[i]) continue;
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < n; k++) {
        if (j == i || k == i || j == k) continue;
        if (angle(mainasu(p[i], p[j]), mainasu(p[i], p[k])) <= 0) {
          ok[j] = false;
          ok[k] = false;
        } else {
          flag = false;
        }
      }
    }
    if (flag) ans.push_back(i + 1);
  }
  cout << ans.size() << endl;
  for (auto anss : ans) {
    cout << anss << endl;
  }
  return 0;
}

```