#include <bits/stdc++.h>
using namespace std;
template <typename A, typename B>
inline void chmin(A &a, B b) {
  if (a > b) a = b;
}
template <typename A, typename B>
inline void chmax(A &a, B b) {
  if (a < b) a = b;
}
long long N;
vector<long long> A[1111];
signed main() {
  cin >> N;
  for (long long i = 0; i < (N); i++) {
    A[i].resize(5);
    for (long long j = 0; j < (5); j++) cin >> A[i][j];
  }
  vector<long long> ans;
  for (long long i = 0; i < (N); i++) {
    vector<vector<long long> > v;
    for (long long j = 0; j < (N); j++) {
      if (i == j) continue;
      vector<long long> x(5);
      for (long long k = 0; k < (5); k++) x[k] = A[j][k] - A[i][k];
      v.push_back(x);
      if (v.size() > 40) break;
    }
    if (v.size() > 40) continue;
    bool flag = true;
    for (long long k = 0; k < (v.size()); k++)
      for (long long l = 0; l < (v.size()); l++) {
        if (k == l) continue;
        long long c = 0;
        for (long long m = 0; m < (5); m++) c += v[k][m] * v[l][m];
        if (c > 0) flag = false;
      }
    if (flag) ans.push_back(i);
  }
  cout << ans.size() << endl;
  for (long long i = 0; i < (ans.size()); i++) cout << ans[i] + 1 << endl;
  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;
template <typename A, typename B>
inline void chmin(A &a, B b) {
  if (a > b) a = b;
}
template <typename A, typename B>
inline void chmax(A &a, B b) {
  if (a < b) a = b;
}
long long N;
vector<long long> A[1111];
signed main() {
  cin >> N;
  for (long long i = 0; i < (N); i++) {
    A[i].resize(5);
    for (long long j = 0; j < (5); j++) cin >> A[i][j];
  }
  vector<long long> ans;
  for (long long i = 0; i < (N); i++) {
    vector<vector<long long> > v;
    for (long long j = 0; j < (N); j++) {
      if (i == j) continue;
      vector<long long> x(5);
      for (long long k = 0; k < (5); k++) x[k] = A[j][k] - A[i][k];
      v.push_back(x);
      if (v.size() > 40) break;
    }
    if (v.size() > 40) continue;
    bool flag = true;
    for (long long k = 0; k < (v.size()); k++)
      for (long long l = 0; l < (v.size()); l++) {
        if (k == l) continue;
        long long c = 0;
        for (long long m = 0; m < (5); m++) c += v[k][m] * v[l][m];
        if (c > 0) flag = false;
      }
    if (flag) ans.push_back(i);
  }
  cout << ans.size() << endl;
  for (long long i = 0; i < (ans.size()); i++) cout << ans[i] + 1 << endl;
  return 0;
}

```