#include <bits/stdc++.h>
using namespace std;
int X[1010][5];
bool bad[1010];
int cross(int a, int b, int c) {
  int re = 0;
  for (int i = 0; i < 5; i++) re += (X[b][i] - X[a][i]) * (X[c][i] - X[a][i]);
  return re;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < 5; j++) cin >> X[i][j];
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++)
      for (int k = 1; k <= n; k++) {
        if (i != j && j != k && k != i && cross(i, j, k) > 0)
          bad[i] = 1, k = n, j = n;
      }
  }
  int num = 0;
  for (int i = 1; i <= n; !bad[i] ? num++ : 1, i++)
    ;
  cout << num << endl;
  for (int i = 1; i <= n; !bad[i] ? printf("%d ", i) : 1, i++)
    ;
  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;
int X[1010][5];
bool bad[1010];
int cross(int a, int b, int c) {
  int re = 0;
  for (int i = 0; i < 5; i++) re += (X[b][i] - X[a][i]) * (X[c][i] - X[a][i]);
  return re;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < 5; j++) cin >> X[i][j];
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++)
      for (int k = 1; k <= n; k++) {
        if (i != j && j != k && k != i && cross(i, j, k) > 0)
          bad[i] = 1, k = n, j = n;
      }
  }
  int num = 0;
  for (int i = 1; i <= n; !bad[i] ? num++ : 1, i++)
    ;
  cout << num << endl;
  for (int i = 1; i <= n; !bad[i] ? printf("%d ", i) : 1, i++)
    ;
  return 0;
}

```