#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
const int N = 3000;
const int R = 5;
double a[N][R];
double dot(const double* a, const double* b) {
  double res = 0;
  for (int i = 0; i < R; ++i) {
    res += a[i] * b[i];
  }
  return res;
}
double len(const double* a) { return sqrt(dot(a, a)); }
double angle(const double* a, const double* b) {
  return acos(dot(a, b) / (len(a) * len(b)));
}
int main() {
  ios_base::sync_with_stdio(0);
  const int B = (1 << 5);
  int n;
  cin >> n;
  if (n > B) {
    cout << 0 << endl;
    return 0;
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < R; ++j) {
      cin >> a[i][j];
    }
  }
  vector<int> res;
  for (int i = 0; i < n; ++i) {
    bool fail = 0;
    for (int j = 0; j < n; ++j) {
      if (j == i) continue;
      for (int k = j + 1; k < n; ++k) {
        if (k == i) continue;
        double x[R], y[R];
        for (int p = 0; p < R; ++p) {
          x[p] = a[j][p] - a[i][p];
          y[p] = a[k][p] - a[i][p];
        }
        auto ang = angle(x, y);
        if (2 * ang < 3.14159265358979323846) {
          fail = 1;
          break;
        }
      }
      if (fail) break;
    }
    if (!fail) {
      res.push_back(i);
    }
  }
  cout << res.size() << endl;
  for (auto e : res) {
    cout << e + 1 << endl;
  }
  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>
using namespace std;
const int INF = 1e9;
const int N = 3000;
const int R = 5;
double a[N][R];
double dot(const double* a, const double* b) {
  double res = 0;
  for (int i = 0; i < R; ++i) {
    res += a[i] * b[i];
  }
  return res;
}
double len(const double* a) { return sqrt(dot(a, a)); }
double angle(const double* a, const double* b) {
  return acos(dot(a, b) / (len(a) * len(b)));
}
int main() {
  ios_base::sync_with_stdio(0);
  const int B = (1 << 5);
  int n;
  cin >> n;
  if (n > B) {
    cout << 0 << endl;
    return 0;
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < R; ++j) {
      cin >> a[i][j];
    }
  }
  vector<int> res;
  for (int i = 0; i < n; ++i) {
    bool fail = 0;
    for (int j = 0; j < n; ++j) {
      if (j == i) continue;
      for (int k = j + 1; k < n; ++k) {
        if (k == i) continue;
        double x[R], y[R];
        for (int p = 0; p < R; ++p) {
          x[p] = a[j][p] - a[i][p];
          y[p] = a[k][p] - a[i][p];
        }
        auto ang = angle(x, y);
        if (2 * ang < 3.14159265358979323846) {
          fail = 1;
          break;
        }
      }
      if (fail) break;
    }
    if (!fail) {
      res.push_back(i);
    }
  }
  cout << res.size() << endl;
  for (auto e : res) {
    cout << e + 1 << endl;
  }
  return 0;
}

```