#include <bits/stdc++.h>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
array<int, 5> operator-(const array<int, 5> &a, const array<int, 5> &b) {
  array<int, 5> c;
  for (int i = 0; i < 5; i++) c[i] = a[i] - b[i];
  return c;
}
int operator*(const array<int, 5> &a, const array<int, 5> &b) {
  int res = 0;
  for (int i = 0; i < 5; i++) {
    res += a[i] * b[i];
  }
  return res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<array<int, 5> > A(n);
  for (int i(0); i < n; i++) {
    for (int j = 0; j < 5; j++) {
      cin >> A[i][j];
    }
  }
  bitset<1010> huh;
  vector<vector<array<int, 5> > > uh(n, vector<array<int, 5> >(n));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      uh[i][j] = A[i] - A[j];
    }
  }
  huh.reset();
  for (int i = 0; i < n; i++) {
    huh[i] = 1;
  }
  array<int, 5> a, b, c;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      for (int k = j + 1; k < n; k++) {
        if (uh[i][j] * uh[i][k] > 0) huh[i] = 0;
        if (uh[i][j] * uh[j][k] < 0) huh[j] = 0;
        if (uh[i][k] * uh[j][k] > 0) huh[k] = 0;
      }
    }
  }
  int cnt = 0;
  for (int i = 0; i < n; i++)
    if (huh[i] == 1) cnt++;
  cout << cnt << '\n';
  for (int i = 0; i < n; i++) {
    if (huh[i] == 1) cout << i + 1 << "\n";
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
array<int, 5> operator-(const array<int, 5> &a, const array<int, 5> &b) {
  array<int, 5> c;
  for (int i = 0; i < 5; i++) c[i] = a[i] - b[i];
  return c;
}
int operator*(const array<int, 5> &a, const array<int, 5> &b) {
  int res = 0;
  for (int i = 0; i < 5; i++) {
    res += a[i] * b[i];
  }
  return res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<array<int, 5> > A(n);
  for (int i(0); i < n; i++) {
    for (int j = 0; j < 5; j++) {
      cin >> A[i][j];
    }
  }
  bitset<1010> huh;
  vector<vector<array<int, 5> > > uh(n, vector<array<int, 5> >(n));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      uh[i][j] = A[i] - A[j];
    }
  }
  huh.reset();
  for (int i = 0; i < n; i++) {
    huh[i] = 1;
  }
  array<int, 5> a, b, c;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      for (int k = j + 1; k < n; k++) {
        if (uh[i][j] * uh[i][k] > 0) huh[i] = 0;
        if (uh[i][j] * uh[j][k] < 0) huh[j] = 0;
        if (uh[i][k] * uh[j][k] > 0) huh[k] = 0;
      }
    }
  }
  int cnt = 0;
  for (int i = 0; i < n; i++)
    if (huh[i] == 1) cnt++;
  cout << cnt << '\n';
  for (int i = 0; i < n; i++) {
    if (huh[i] == 1) cout << i + 1 << "\n";
  }
  return 0;
}

```