#include <bits/stdc++.h>
inline int two(int n) { return 1 << n; }
inline int test(int n, int b) { return (n >> b) & 1; }
inline void set_bit(int& n, int b) { n |= two(b); }
inline void unset_bit(int& n, int b) { n &= ~two(b); }
inline int last_bit(int n) { return n & (-n); }
inline int ones(int n) {
  int res = 0;
  while (n && ++res) n -= n & (-n);
  return res;
}
long long int gcd(long long int a, long long int b) {
  return (a ? gcd(b % a, a) : b);
}
long long int modPow(long long int a, long long int b, long long int MOD) {
  long long int x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y) % MOD;
    }
    b /= 2;
    y = (y * y) % MOD;
  }
  return x;
}
long long int modInverse(long long int a, long long int p) {
  return modPow(a, p - 2, p);
}
using namespace std;
const int N = 1e3 + 3;
int arr[N][5];
vector<int> ans;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, i, j, k, l;
  cin >> n;
  if (n > 100) {
    cout << 0;
    return 0;
  }
  for (i = 0; i < (n); i++) {
    for (j = 0; j < (5); j++) cin >> arr[i][j];
  }
  for (i = 0; i < (n); i++) {
    bool flag = false;
    for (j = 0; j < (n); j++) {
      for (k = 0; k < (n); k++) {
        if (i != j && i != k && k != j) {
          long long int sum = 0;
          for (l = 0; l < (5); l++)
            sum += ((arr[i][l] - arr[j][l]) * (arr[i][l] - arr[k][l]));
          if (sum > 0) flag = true;
        }
      }
    }
    if (!flag) ans.push_back(i + 1);
  }
  cout << ans.size() << "\n";
  for (i = 0; i < (ans.size()); i++) cout << ans[i] << " ";
  return 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>
inline int two(int n) { return 1 << n; }
inline int test(int n, int b) { return (n >> b) & 1; }
inline void set_bit(int& n, int b) { n |= two(b); }
inline void unset_bit(int& n, int b) { n &= ~two(b); }
inline int last_bit(int n) { return n & (-n); }
inline int ones(int n) {
  int res = 0;
  while (n && ++res) n -= n & (-n);
  return res;
}
long long int gcd(long long int a, long long int b) {
  return (a ? gcd(b % a, a) : b);
}
long long int modPow(long long int a, long long int b, long long int MOD) {
  long long int x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y) % MOD;
    }
    b /= 2;
    y = (y * y) % MOD;
  }
  return x;
}
long long int modInverse(long long int a, long long int p) {
  return modPow(a, p - 2, p);
}
using namespace std;
const int N = 1e3 + 3;
int arr[N][5];
vector<int> ans;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, i, j, k, l;
  cin >> n;
  if (n > 100) {
    cout << 0;
    return 0;
  }
  for (i = 0; i < (n); i++) {
    for (j = 0; j < (5); j++) cin >> arr[i][j];
  }
  for (i = 0; i < (n); i++) {
    bool flag = false;
    for (j = 0; j < (n); j++) {
      for (k = 0; k < (n); k++) {
        if (i != j && i != k && k != j) {
          long long int sum = 0;
          for (l = 0; l < (5); l++)
            sum += ((arr[i][l] - arr[j][l]) * (arr[i][l] - arr[k][l]));
          if (sum > 0) flag = true;
        }
      }
    }
    if (!flag) ans.push_back(i + 1);
  }
  cout << ans.size() << "\n";
  for (i = 0; i < (ans.size()); i++) cout << ans[i] << " ";
  return 0;
}

```