#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
const ll MOD = 998244353;
ll binpow(ll a, ll b) {
  ll res = 1;
  while (b) {
    if (b & 1) res = (res * a) % MOD;
    a = (a * a) % MOD;
    b >>= 1;
  }
  return res;
}
ll modInv(ll a) { return binpow(a, MOD - 2); }
const double PI = acos(-1);
const int INF = 0x3f3f3f3f;
const int NINF = 0xc0c0c0c0;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const ll NINFLL = 0xc0c0c0c0c0c0c0c0;
const int mxN = 1001;
array<int, 5> arr[mxN];
int n;
int main() {
  mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
  cin.tie(0)->sync_with_stdio(0);
  cin >> n;
  if (n >= 50) {
    cout << "0\n";
    return 0;
  }
  for (int i = (0); i < (n); ++i) {
    cin >> arr[i][0] >> arr[i][1] >> arr[i][2] >> arr[i][3] >> arr[i][4];
  }
  vector<int> ans;
  for (int i = (0); i < (n); ++i) {
    bool ok = true;
    int num = 0;
    for (int j = (0); j < (n); ++j) {
      for (int k = (0); k < (n); ++k) {
        if (i == j || j == k || i == k) continue;
        num = (arr[j][0] - arr[i][0]) * (arr[k][0] - arr[i][0]) +
              (arr[j][1] - arr[i][1]) * (arr[k][1] - arr[i][1]) +
              (arr[j][2] - arr[i][2]) * (arr[k][2] - arr[i][2]) +
              (arr[j][3] - arr[i][3]) * (arr[k][3] - arr[i][3]) +
              (arr[j][4] - arr[i][4]) * (arr[k][4] - arr[i][4]);
        if (num > 0) ok = false;
      }
    }
    if (ok) ans.push_back(i);
  }
  cout << ans.size() << "\n";
  for (int j : ans) {
    cout << (j + 1) << "\n";
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
using ll = long long;
using ld = long double;
const ll MOD = 998244353;
ll binpow(ll a, ll b) {
  ll res = 1;
  while (b) {
    if (b & 1) res = (res * a) % MOD;
    a = (a * a) % MOD;
    b >>= 1;
  }
  return res;
}
ll modInv(ll a) { return binpow(a, MOD - 2); }
const double PI = acos(-1);
const int INF = 0x3f3f3f3f;
const int NINF = 0xc0c0c0c0;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const ll NINFLL = 0xc0c0c0c0c0c0c0c0;
const int mxN = 1001;
array<int, 5> arr[mxN];
int n;
int main() {
  mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
  cin.tie(0)->sync_with_stdio(0);
  cin >> n;
  if (n >= 50) {
    cout << "0\n";
    return 0;
  }
  for (int i = (0); i < (n); ++i) {
    cin >> arr[i][0] >> arr[i][1] >> arr[i][2] >> arr[i][3] >> arr[i][4];
  }
  vector<int> ans;
  for (int i = (0); i < (n); ++i) {
    bool ok = true;
    int num = 0;
    for (int j = (0); j < (n); ++j) {
      for (int k = (0); k < (n); ++k) {
        if (i == j || j == k || i == k) continue;
        num = (arr[j][0] - arr[i][0]) * (arr[k][0] - arr[i][0]) +
              (arr[j][1] - arr[i][1]) * (arr[k][1] - arr[i][1]) +
              (arr[j][2] - arr[i][2]) * (arr[k][2] - arr[i][2]) +
              (arr[j][3] - arr[i][3]) * (arr[k][3] - arr[i][3]) +
              (arr[j][4] - arr[i][4]) * (arr[k][4] - arr[i][4]);
        if (num > 0) ok = false;
      }
    }
    if (ok) ans.push_back(i);
  }
  cout << ans.size() << "\n";
  for (int j : ans) {
    cout << (j + 1) << "\n";
  }
  return 0;
}

```