#include <bits/stdc++.h>
using namespace std;
int64_t mod = 1e9 + 7;
int64_t mod1 = 1e9 + 6;
int64_t power(int64_t a, int64_t b) {
  if (b == 0)
    return 1;
  else if (b % 2 == 0)
    return power((((a % mod) * (a % mod)) % mod), b / 2) % mod;
  else
    return ((a % mod) * (power((((a % mod) * (a % mod)) % mod), b / 2) % mod)) %
           mod;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int64_t n;
  cin >> n;
  int64_t a[n + 1], b[n + 1], c[n + 1], d[n + 1], e[n + 1];
  for (int64_t i = 1; i <= n; ++i) cin >> a[i] >> b[i] >> c[i] >> d[i] >> e[i];
  if (n <= 2) {
    cout << n << '\n';
    for (int64_t i = 1; i <= n; ++i) cout << i << '\n';
    return 0;
  }
  bool vis[n + 1];
  memset(vis, false, sizeof(vis));
  for (int64_t i = 1; i <= n; ++i) {
    if (vis[i]) continue;
    int64_t ch = 0;
    for (int64_t j = 1; j <= n; ++j) {
      if (i == j) continue;
      for (int64_t k = j + 1; k <= n; ++k) {
        if (i == k or k == j) continue;
        int64_t s =
            (a[j] - a[i]) * (a[k] - a[i]) + (b[j] - b[i]) * (b[k] - b[i]) +
            (c[j] - c[i]) * (c[k] - c[i]) + (d[j] - d[i]) * (d[k] - d[i]) +
            (e[j] - e[i]) * (e[k] - e[i]);
        if (s > 0) {
          ch = 1;
          break;
        }
      }
      if (ch) break;
    }
    if (ch == 0) {
      cout << 1 << '\n';
      cout << i;
      return 0;
    }
  }
  cout << 0;
}

### Prompt

Please formulate 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;
int64_t mod = 1e9 + 7;
int64_t mod1 = 1e9 + 6;
int64_t power(int64_t a, int64_t b) {
  if (b == 0)
    return 1;
  else if (b % 2 == 0)
    return power((((a % mod) * (a % mod)) % mod), b / 2) % mod;
  else
    return ((a % mod) * (power((((a % mod) * (a % mod)) % mod), b / 2) % mod)) %
           mod;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int64_t n;
  cin >> n;
  int64_t a[n + 1], b[n + 1], c[n + 1], d[n + 1], e[n + 1];
  for (int64_t i = 1; i <= n; ++i) cin >> a[i] >> b[i] >> c[i] >> d[i] >> e[i];
  if (n <= 2) {
    cout << n << '\n';
    for (int64_t i = 1; i <= n; ++i) cout << i << '\n';
    return 0;
  }
  bool vis[n + 1];
  memset(vis, false, sizeof(vis));
  for (int64_t i = 1; i <= n; ++i) {
    if (vis[i]) continue;
    int64_t ch = 0;
    for (int64_t j = 1; j <= n; ++j) {
      if (i == j) continue;
      for (int64_t k = j + 1; k <= n; ++k) {
        if (i == k or k == j) continue;
        int64_t s =
            (a[j] - a[i]) * (a[k] - a[i]) + (b[j] - b[i]) * (b[k] - b[i]) +
            (c[j] - c[i]) * (c[k] - c[i]) + (d[j] - d[i]) * (d[k] - d[i]) +
            (e[j] - e[i]) * (e[k] - e[i]);
        if (s > 0) {
          ch = 1;
          break;
        }
      }
      if (ch) break;
    }
    if (ch == 0) {
      cout << 1 << '\n';
      cout << i;
      return 0;
    }
  }
  cout << 0;
}

```