#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0), EPS = 1e-13;
bool isacute(vector<double> x, vector<double> y) {
  double xy = 0;
  for (int i = 0; i < 5; ++i) {
    xy += x[i] * y[i];
  }
  return xy > EPS;
}
int main() {
  int n;
  cin >> n;
  vector<vector<double> > a(n, vector<double>(5));
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < 5; ++j) {
      scanf("%lf", &a[i][j]);
    }
  vector<int> ok(n, 1);
  for (int i = 0; i < n; ++i) {
    if (!ok[i]) continue;
    for (int j = 0; j < n; ++j) {
      if (j == i) continue;
      for (int k = 0; k < j; ++k) {
        if (k == i) continue;
        vector<double> ab(5), ac(5);
        for (int p = 0; p < 5; ++p) {
          ab[p] = a[j][p] - a[i][p];
          ac[p] = a[k][p] - a[i][p];
        }
        if (isacute(ab, ac)) {
          ok[i] = 0;
          j = n;
          break;
        } else {
          ok[j] = 0;
          ok[k] = 0;
        }
      }
    }
  }
  int num = 0;
  for (int i = 0; i < n; ++i) num += ok[i];
  cout << num << endl;
  for (int i = 0; i < n; ++i)
    if (ok[i]) printf("%d\n", i + 1);
  return 0;
}

### Prompt

Please formulate 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;
const double pi = acos(-1.0), EPS = 1e-13;
bool isacute(vector<double> x, vector<double> y) {
  double xy = 0;
  for (int i = 0; i < 5; ++i) {
    xy += x[i] * y[i];
  }
  return xy > EPS;
}
int main() {
  int n;
  cin >> n;
  vector<vector<double> > a(n, vector<double>(5));
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < 5; ++j) {
      scanf("%lf", &a[i][j]);
    }
  vector<int> ok(n, 1);
  for (int i = 0; i < n; ++i) {
    if (!ok[i]) continue;
    for (int j = 0; j < n; ++j) {
      if (j == i) continue;
      for (int k = 0; k < j; ++k) {
        if (k == i) continue;
        vector<double> ab(5), ac(5);
        for (int p = 0; p < 5; ++p) {
          ab[p] = a[j][p] - a[i][p];
          ac[p] = a[k][p] - a[i][p];
        }
        if (isacute(ab, ac)) {
          ok[i] = 0;
          j = n;
          break;
        } else {
          ok[j] = 0;
          ok[k] = 0;
        }
      }
    }
  }
  int num = 0;
  for (int i = 0; i < n; ++i) num += ok[i];
  cout << num << endl;
  for (int i = 0; i < n; ++i)
    if (ok[i]) printf("%d\n", i + 1);
  return 0;
}

```