#include <bits/stdc++.h>
using namespace std;
int n;
int pts[1005][5];
bool bad(int i) {
  int x[5];
  for (int j = 0; j < n; ++j) {
    if (j == i) continue;
    for (int t = 0; t < 5; ++t) {
      x[t] = pts[i][t] - pts[j][t];
    }
    for (int k = j + 1; k < n; ++k) {
      if (k == i) continue;
      int res = 0;
      for (int t = 0; t < 5; ++t) {
        res += (x[t]) * (pts[i][t] - pts[k][t]);
      }
      if (res > 0) return true;
    }
  }
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> pts[i][0] >> pts[i][1] >> pts[i][2] >> pts[i][3] >> pts[i][4];
  }
  int cnt = 0;
  vector<int> res;
  res.reserve(1005);
  for (int i = 0; i < n; ++i) {
    if (bad(i)) continue;
    ++cnt;
    res.push_back(i + 1);
  }
  cout << cnt << endl;
  for (int i = 0; i < cnt; ++i) {
    cout << res[i] << endl;
  }
  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;
int n;
int pts[1005][5];
bool bad(int i) {
  int x[5];
  for (int j = 0; j < n; ++j) {
    if (j == i) continue;
    for (int t = 0; t < 5; ++t) {
      x[t] = pts[i][t] - pts[j][t];
    }
    for (int k = j + 1; k < n; ++k) {
      if (k == i) continue;
      int res = 0;
      for (int t = 0; t < 5; ++t) {
        res += (x[t]) * (pts[i][t] - pts[k][t]);
      }
      if (res > 0) return true;
    }
  }
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> pts[i][0] >> pts[i][1] >> pts[i][2] >> pts[i][3] >> pts[i][4];
  }
  int cnt = 0;
  vector<int> res;
  res.reserve(1005);
  for (int i = 0; i < n; ++i) {
    if (bad(i)) continue;
    ++cnt;
    res.push_back(i + 1);
  }
  cout << cnt << endl;
  for (int i = 0; i < cnt; ++i) {
    cout << res[i] << endl;
  }
  return 0;
}

```