#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3 + 10;
const int maxint = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int n, p[maxn][5], a[maxn], ans;
int t[maxn], f[maxn][maxn];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 5; j++) {
      cin >> p[i][j];
    }
  }
  for (int x = 1; x <= n; x++) {
    for (int j = 0; j < 5; j++) {
      t[x] += p[x][j] * p[x][j];
      for (int y = 1; y <= n; y++) {
        f[x][y] += p[x][j] * p[y][j];
      }
    }
  }
  ans = n;
  for (int x = 1; x <= n; x++) {
    for (int y = 1; y < n && !a[x]; y++) {
      for (int z = y + 1; z <= n && !a[x]; z++) {
        if (y != x && z != x) {
          if (t[x] + f[y][z] - f[x][y] - f[x][z] > 0) {
            a[x] = 1;
            ans--;
          }
        }
      }
    }
  }
  cout << ans << "\n";
  for (int i = 1; i <= n; i++) {
    if (!a[i]) cout << i << "\n";
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
const int maxn = 1e3 + 10;
const int maxint = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int n, p[maxn][5], a[maxn], ans;
int t[maxn], f[maxn][maxn];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 5; j++) {
      cin >> p[i][j];
    }
  }
  for (int x = 1; x <= n; x++) {
    for (int j = 0; j < 5; j++) {
      t[x] += p[x][j] * p[x][j];
      for (int y = 1; y <= n; y++) {
        f[x][y] += p[x][j] * p[y][j];
      }
    }
  }
  ans = n;
  for (int x = 1; x <= n; x++) {
    for (int y = 1; y < n && !a[x]; y++) {
      for (int z = y + 1; z <= n && !a[x]; z++) {
        if (y != x && z != x) {
          if (t[x] + f[y][z] - f[x][y] - f[x][z] > 0) {
            a[x] = 1;
            ans--;
          }
        }
      }
    }
  }
  cout << ans << "\n";
  for (int i = 1; i <= n; i++) {
    if (!a[i]) cout << i << "\n";
  }
  return 0;
}

```