#include <bits/stdc++.h>
using namespace std;
template <class T>
bool chmin(T &a, T b) {
  return b < a ? (a = b, true) : false;
}
template <class T>
bool chmax(T &a, T b) {
  return b > a ? (a = b, true) : false;
}
void kout() { cerr << endl; }
template <class T, class... U>
void kout(T a, U... b) {
  cerr << a << ' ', kout(b...);
}
template <class T>
void pary(T L, T R) {
  while (L != R) cerr << *L << " \n"[++L == R];
}
const int MAXN = 1000;
const double PI = 3.1415926535897932384626;
int n;
int arr[MAXN][5];
vector<int> ans;
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n;
  if (n > 50) {
    cout << 0 << "\n";
    return 0;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      cin >> arr[i][j];
    }
  }
  for (int i = 0; i < n; i++) {
    bool flag = 1;
    for (int a = 0; a < n; a++) {
      if (a == i) continue;
      for (int b = 0; b < n; b++) {
        if (a == b || b == i) continue;
        double A = 0, B = 0, C = 0;
        for (int k = 0; k < 5; k++) {
          A += (arr[a][k] - arr[i][k]) * (arr[a][k] - arr[i][k]);
          B += (arr[b][k] - arr[i][k]) * (arr[b][k] - arr[i][k]);
          C += (arr[a][k] - arr[i][k]) * (arr[b][k] - arr[i][k]);
        }
        if (acos(C / sqrt(A * B)) * 180 / PI < 90) flag = 0;
      }
    }
    if (flag) ans.push_back(i + 1);
  }
  cout << ans.size() << "\n";
  for (int i : ans) cout << i << "\n";
}

### 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>
using namespace std;
template <class T>
bool chmin(T &a, T b) {
  return b < a ? (a = b, true) : false;
}
template <class T>
bool chmax(T &a, T b) {
  return b > a ? (a = b, true) : false;
}
void kout() { cerr << endl; }
template <class T, class... U>
void kout(T a, U... b) {
  cerr << a << ' ', kout(b...);
}
template <class T>
void pary(T L, T R) {
  while (L != R) cerr << *L << " \n"[++L == R];
}
const int MAXN = 1000;
const double PI = 3.1415926535897932384626;
int n;
int arr[MAXN][5];
vector<int> ans;
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n;
  if (n > 50) {
    cout << 0 << "\n";
    return 0;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      cin >> arr[i][j];
    }
  }
  for (int i = 0; i < n; i++) {
    bool flag = 1;
    for (int a = 0; a < n; a++) {
      if (a == i) continue;
      for (int b = 0; b < n; b++) {
        if (a == b || b == i) continue;
        double A = 0, B = 0, C = 0;
        for (int k = 0; k < 5; k++) {
          A += (arr[a][k] - arr[i][k]) * (arr[a][k] - arr[i][k]);
          B += (arr[b][k] - arr[i][k]) * (arr[b][k] - arr[i][k]);
          C += (arr[a][k] - arr[i][k]) * (arr[b][k] - arr[i][k]);
        }
        if (acos(C / sqrt(A * B)) * 180 / PI < 90) flag = 0;
      }
    }
    if (flag) ans.push_back(i + 1);
  }
  cout << ans.size() << "\n";
  for (int i : ans) cout << i << "\n";
}

```