#include <bits/stdc++.h>
using namespace std;
struct point {
  int x[5];
};
int N;
point P[1005];
vector<int> ans;
int dot(point a, point b) {
  int ret = 0;
  for (int i = 0; i < 5; i++) ret += a.x[i] * b.x[i];
  return ret;
}
point subtract(point a, point b) {
  point c;
  for (int i = 0; i < 5; i++) c.x[i] = a.x[i] - b.x[i];
  return c;
}
bool check(int cur) {
  for (int i = 0; i < N; i++)
    for (int j = 0; j < N; j++)
      if (i != cur && j != cur && i != j) {
        point a = subtract(P[i], P[cur]), b = subtract(P[j], P[cur]);
        if (dot(a, b) > 0) return false;
      }
  return true;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> N;
  for (int i = 0; i < N; i++)
    for (int j = 0; j < 5; j++) cin >> P[i].x[j];
  for (int i = 0; i < N; i++)
    if (check(i)) ans.push_back(i);
  cout << ans.size() << "\n";
  if (!ans.empty()) {
    for (int i = 0; i < ans.size(); i++) cout << ans[i] + 1 << " ";
    cout << "\n";
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution 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;
struct point {
  int x[5];
};
int N;
point P[1005];
vector<int> ans;
int dot(point a, point b) {
  int ret = 0;
  for (int i = 0; i < 5; i++) ret += a.x[i] * b.x[i];
  return ret;
}
point subtract(point a, point b) {
  point c;
  for (int i = 0; i < 5; i++) c.x[i] = a.x[i] - b.x[i];
  return c;
}
bool check(int cur) {
  for (int i = 0; i < N; i++)
    for (int j = 0; j < N; j++)
      if (i != cur && j != cur && i != j) {
        point a = subtract(P[i], P[cur]), b = subtract(P[j], P[cur]);
        if (dot(a, b) > 0) return false;
      }
  return true;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> N;
  for (int i = 0; i < N; i++)
    for (int j = 0; j < 5; j++) cin >> P[i].x[j];
  for (int i = 0; i < N; i++)
    if (check(i)) ans.push_back(i);
  cout << ans.size() << "\n";
  if (!ans.empty()) {
    for (int i = 0; i < ans.size(); i++) cout << ans[i] + 1 << " ";
    cout << "\n";
  }
  return 0;
}

```