#include <bits/stdc++.h>
using namespace std;
void RI() {}
template <typename... T>
void RI(int& head, T&... tail) {
  scanf("%d", &head);
  RI(tail...);
}
void RLL() {}
template <typename... T>
void RLL(long long& head, T&... tail) {
  scanf("%lld", &head);
  RLL(tail...);
}
void print() { putchar('\n'); }
template <typename T, typename... U>
void print(T head, U... tail) {
  cout << head << " ";
  print(tail...);
}
const int MOD = 1000000007;
int a[1005][5];
int main() {
  int n;
  RI(n);
  for (int i = 0; i < (int)(n); i++) {
    for (int d = 0; d < (int)(5); d++) {
      RI(a[i][d]);
    }
  }
  if (n >= 65) {
    print("0\n");
    return 0;
  }
  unordered_set<int> bad_idxes;
  for (int i = 0; i < (int)(n); i++) {
    bool bad = false;
    for (int j = 0; j < (int)(n); j++) {
      for (int k = j + 1; k < (int)(n); k++) {
        int dot = 0;
        for (int d = 0; d < (int)(5); d++) {
          dot += (a[j][d] - a[i][d]) * (a[k][d] - a[i][d]);
        }
        if (dot > 0) {
          bad = true;
          break;
        }
      }
      if (bad) {
        break;
      }
    }
    if (bad) {
      bad_idxes.insert(i + 1);
    }
  }
  printf("%d\n", n - (int)bad_idxes.size());
  for (int i = 1; i < (int)(n + 1); i++) {
    if (bad_idxes.find(i) == bad_idxes.end()) {
      printf("%d\n", i);
    }
  }
  return 0;
}

### Prompt

Generate 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;
void RI() {}
template <typename... T>
void RI(int& head, T&... tail) {
  scanf("%d", &head);
  RI(tail...);
}
void RLL() {}
template <typename... T>
void RLL(long long& head, T&... tail) {
  scanf("%lld", &head);
  RLL(tail...);
}
void print() { putchar('\n'); }
template <typename T, typename... U>
void print(T head, U... tail) {
  cout << head << " ";
  print(tail...);
}
const int MOD = 1000000007;
int a[1005][5];
int main() {
  int n;
  RI(n);
  for (int i = 0; i < (int)(n); i++) {
    for (int d = 0; d < (int)(5); d++) {
      RI(a[i][d]);
    }
  }
  if (n >= 65) {
    print("0\n");
    return 0;
  }
  unordered_set<int> bad_idxes;
  for (int i = 0; i < (int)(n); i++) {
    bool bad = false;
    for (int j = 0; j < (int)(n); j++) {
      for (int k = j + 1; k < (int)(n); k++) {
        int dot = 0;
        for (int d = 0; d < (int)(5); d++) {
          dot += (a[j][d] - a[i][d]) * (a[k][d] - a[i][d]);
        }
        if (dot > 0) {
          bad = true;
          break;
        }
      }
      if (bad) {
        break;
      }
    }
    if (bad) {
      bad_idxes.insert(i + 1);
    }
  }
  printf("%d\n", n - (int)bad_idxes.size());
  for (int i = 1; i < (int)(n + 1); i++) {
    if (bad_idxes.find(i) == bad_idxes.end()) {
      printf("%d\n", i);
    }
  }
  return 0;
}

```