#include <bits/stdc++.h>
using namespace std;
unordered_map<long long int, int> s;
unordered_map<long long int, long long int> po;
vector<long long int> b;
void SieveOfEratosthenes(int n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * 2; i <= n; i += p) {
        if (prime[i] == true) {
          prime[i] = false;
          b.emplace_back(p);
        }
      }
    }
  }
}
int main() {
  long long int n;
  cin >> n;
  long long int a[n][5];
  for (long long int i = 0; i < n; i++) {
    for (long long int j = 0; j < 5; j++) {
      cin >> a[i][j];
    }
  }
  if (n > 100) {
    cout << "0";
    return 0;
  }
  vector<long long int> v;
  for (long long int i = 0; i < n; i++) {
    int w = 0;
    for (long long int j = 0; j < n; j++) {
      if (j == i) {
        continue;
      }
      int r = 0;
      for (long long int k = 0; k < n; k++) {
        if (k == i || k == j) {
          continue;
        }
        long long int ans = 0;
        for (long long int y = 0; y < 5; y++) {
          ans += ((a[k][y] - a[i][y]) * (a[j][y] - a[i][y]));
        }
        if (ans > 0) {
          r = 1;
          break;
        }
      }
      if (r == 1) {
        w++;
        break;
      }
    }
    if (w == 0) {
      v.emplace_back(i + 1);
    }
  }
  cout << v.size() << "\n";
  for (long long int i = 0; i < v.size(); i++) {
    cout << v[i] << " ";
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
unordered_map<long long int, int> s;
unordered_map<long long int, long long int> po;
vector<long long int> b;
void SieveOfEratosthenes(int n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * 2; i <= n; i += p) {
        if (prime[i] == true) {
          prime[i] = false;
          b.emplace_back(p);
        }
      }
    }
  }
}
int main() {
  long long int n;
  cin >> n;
  long long int a[n][5];
  for (long long int i = 0; i < n; i++) {
    for (long long int j = 0; j < 5; j++) {
      cin >> a[i][j];
    }
  }
  if (n > 100) {
    cout << "0";
    return 0;
  }
  vector<long long int> v;
  for (long long int i = 0; i < n; i++) {
    int w = 0;
    for (long long int j = 0; j < n; j++) {
      if (j == i) {
        continue;
      }
      int r = 0;
      for (long long int k = 0; k < n; k++) {
        if (k == i || k == j) {
          continue;
        }
        long long int ans = 0;
        for (long long int y = 0; y < 5; y++) {
          ans += ((a[k][y] - a[i][y]) * (a[j][y] - a[i][y]));
        }
        if (ans > 0) {
          r = 1;
          break;
        }
      }
      if (r == 1) {
        w++;
        break;
      }
    }
    if (w == 0) {
      v.emplace_back(i + 1);
    }
  }
  cout << v.size() << "\n";
  for (long long int i = 0; i < v.size(); i++) {
    cout << v[i] << " ";
  }
  return 0;
}

```