#include <bits/stdc++.h>
using namespace std;
struct point {
  long long a, b, c, d, e;
};
int main() {
  long long n;
  cin >> n;
  multiset<pair<double, pair<int, int> > > ms;
  vector<point> a(n);
  long long i, j;
  for (i = 0; i < n; i++) cin >> a[i].a >> a[i].b >> a[i].c >> a[i].d >> a[i].e;
  for (i = 0; i < n; i++) {
    for (j = i + 1; j < n; j++) {
      double val = (double)pow(a[i].a - a[j].a, (long long)2);
      val += (double)pow(a[i].b - a[j].b, (long long)2);
      val += (double)pow(a[i].c - a[j].c, (long long)2);
      val += (double)pow(a[i].d - a[j].d, (long long)2);
      val += (double)pow(a[i].e - a[j].e, (long long)2);
      val = sqrt(val);
      ms.insert({val, {i, j}});
    }
  }
  vector<int> ans(n);
  long long k = 0;
  for (i = 0; i < n; i++) {
    bool check = true;
    for (auto x = ms.begin(); x != ms.end(); x++) {
      if (x->second.first == i || x->second.second == i)
        continue;
      else {
        long long u = x->second.first;
        long long v = x->second.second;
        long long val = (a[u].a - a[i].a) * (a[v].a - a[i].a);
        val += (a[u].b - a[i].b) * (a[v].b - a[i].b);
        val += (a[u].c - a[i].c) * (a[v].c - a[i].c);
        val += (a[u].d - a[i].d) * (a[v].d - a[i].d);
        val += (a[u].e - a[i].e) * (a[v].e - a[i].e);
        if (val > 0) check = false;
        break;
      }
    }
    if (check) ans[k++] = i;
  }
  cout << k << "\n";
  for (i = 0; i < k; i++) cout << ans[i] + 1 << "\n";
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
struct point {
  long long a, b, c, d, e;
};
int main() {
  long long n;
  cin >> n;
  multiset<pair<double, pair<int, int> > > ms;
  vector<point> a(n);
  long long i, j;
  for (i = 0; i < n; i++) cin >> a[i].a >> a[i].b >> a[i].c >> a[i].d >> a[i].e;
  for (i = 0; i < n; i++) {
    for (j = i + 1; j < n; j++) {
      double val = (double)pow(a[i].a - a[j].a, (long long)2);
      val += (double)pow(a[i].b - a[j].b, (long long)2);
      val += (double)pow(a[i].c - a[j].c, (long long)2);
      val += (double)pow(a[i].d - a[j].d, (long long)2);
      val += (double)pow(a[i].e - a[j].e, (long long)2);
      val = sqrt(val);
      ms.insert({val, {i, j}});
    }
  }
  vector<int> ans(n);
  long long k = 0;
  for (i = 0; i < n; i++) {
    bool check = true;
    for (auto x = ms.begin(); x != ms.end(); x++) {
      if (x->second.first == i || x->second.second == i)
        continue;
      else {
        long long u = x->second.first;
        long long v = x->second.second;
        long long val = (a[u].a - a[i].a) * (a[v].a - a[i].a);
        val += (a[u].b - a[i].b) * (a[v].b - a[i].b);
        val += (a[u].c - a[i].c) * (a[v].c - a[i].c);
        val += (a[u].d - a[i].d) * (a[v].d - a[i].d);
        val += (a[u].e - a[i].e) * (a[v].e - a[i].e);
        if (val > 0) check = false;
        break;
      }
    }
    if (check) ans[k++] = i;
  }
  cout << k << "\n";
  for (i = 0; i < k; i++) cout << ans[i] + 1 << "\n";
  return 0;
}

```