#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e3;
const double PI = atan(1) * 4;
int N, A[MAXN][5];
vector<int> ans;
double ff(int a, int b, int c) {
  double p = 0, q1 = 0, q2 = 0;
  for (int i = 0; i < (int)5; i++) {
    p += (A[b][i] - A[a][i]) * (A[c][i] - A[a][i]);
    q1 += (A[b][i] - A[a][i]) * (A[b][i] - A[a][i]);
    q2 += (A[c][i] - A[a][i]) * (A[c][i] - A[a][i]);
  }
  return acos(p / (sqrt(q1) * sqrt(q2)));
}
bool f(int id) {
  for (int i = 0; i < (int)N; i++)
    for (int j = i + 1; j < (int)N; j++)
      if (i != id && j != id && ff(id, i, j) < PI / 2) return false;
  return true;
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < (int)N; i++)
    for (int j = 0; j < (int)5; j++) scanf("%d", &A[i][j]);
  if (N > 100) {
    printf("0\n");
    return 0;
  }
  for (int i = 0; i < (int)N; i++)
    if (f(i)) ans.push_back(i);
  printf("%d\n", (int)ans.size());
  for (int i = 0; i < (int)ans.size(); i++) printf("%d\n", ans[i] + 1);
}

### Prompt

Create a solution in Cpp for 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 int MAXN = 1e3;
const double PI = atan(1) * 4;
int N, A[MAXN][5];
vector<int> ans;
double ff(int a, int b, int c) {
  double p = 0, q1 = 0, q2 = 0;
  for (int i = 0; i < (int)5; i++) {
    p += (A[b][i] - A[a][i]) * (A[c][i] - A[a][i]);
    q1 += (A[b][i] - A[a][i]) * (A[b][i] - A[a][i]);
    q2 += (A[c][i] - A[a][i]) * (A[c][i] - A[a][i]);
  }
  return acos(p / (sqrt(q1) * sqrt(q2)));
}
bool f(int id) {
  for (int i = 0; i < (int)N; i++)
    for (int j = i + 1; j < (int)N; j++)
      if (i != id && j != id && ff(id, i, j) < PI / 2) return false;
  return true;
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < (int)N; i++)
    for (int j = 0; j < (int)5; j++) scanf("%d", &A[i][j]);
  if (N > 100) {
    printf("0\n");
    return 0;
  }
  for (int i = 0; i < (int)N; i++)
    if (f(i)) ans.push_back(i);
  printf("%d\n", (int)ans.size());
  for (int i = 0; i < (int)ans.size(); i++) printf("%d\n", ans[i] + 1);
}

```