#include <bits/stdc++.h>
using namespace std;
int p[1005][6], n;
bool vis[1005];
void solve() {
  int cnt = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= 5; j++) cin >> p[i][j];
  if (n > 250) {
    puts("0");
    return;
  }
  memset(vis, false, sizeof(vis));
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    for (int j = 1; j <= n; j++) {
      if (j == i) continue;
      for (int k = j + 1; k <= n; k++) {
        if (k == i) continue;
        int val = 0;
        for (int ii = 1; ii <= 5; ii++)
          val += (p[j][ii] - p[i][ii]) * (p[k][ii] - p[i][ii]);
        if (val > 0) {
          vis[i] = true;
          break;
        } else
          vis[j] = vis[k] = true;
      }
      if (vis[i]) break;
    }
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) cnt++;
  cout << cnt << endl;
  for (int i = 1; i <= n; i++)
    if (!vis[i]) cout << i << endl;
  ;
}
int main() {
  while (cin >> n) {
    solve();
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
int p[1005][6], n;
bool vis[1005];
void solve() {
  int cnt = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= 5; j++) cin >> p[i][j];
  if (n > 250) {
    puts("0");
    return;
  }
  memset(vis, false, sizeof(vis));
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    for (int j = 1; j <= n; j++) {
      if (j == i) continue;
      for (int k = j + 1; k <= n; k++) {
        if (k == i) continue;
        int val = 0;
        for (int ii = 1; ii <= 5; ii++)
          val += (p[j][ii] - p[i][ii]) * (p[k][ii] - p[i][ii]);
        if (val > 0) {
          vis[i] = true;
          break;
        } else
          vis[j] = vis[k] = true;
      }
      if (vis[i]) break;
    }
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) cnt++;
  cout << cnt << endl;
  for (int i = 1; i <= n; i++)
    if (!vis[i]) cout << i << endl;
  ;
}
int main() {
  while (cin >> n) {
    solve();
  }
  return 0;
}

```