#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f2f1f0f;
const long long LINF = 1ll * INF * INF;
const int MAX_N = 1e3 + 10;
int N, Nr[MAX_N][5];
int main() {
  cin >> N;
  for (int i = 1; i <= N; i++)
    for (int j = 0; j < 5; j++) scanf("%d", &Nr[i][j]);
  if (N >= 100) {
    puts("0");
    return 0;
  }
  vector<int> Ans;
  for (int i = 1; i <= N; i++) {
    bool isCan = true;
    for (int j = 1; j <= N; j++)
      if (j != i)
        for (int k = j + 1; k <= N; k++)
          if (k != i) {
            int sum = 0;
            for (int b = 0; b < 5; b++)
              sum += (Nr[j][b] - Nr[i][b]) * (Nr[k][b] - Nr[i][b]);
            if (sum > 0) {
              isCan = false;
              goto finish;
            }
          }
  finish:
    if (isCan) Ans.push_back(i);
  }
  printf("%d\n", ((int)(Ans).size()));
  for (int x : Ans) printf("%d\n", x);
  return 0;
}

### Prompt

Please create a solution in cpp 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 int INF = 0x3f2f1f0f;
const long long LINF = 1ll * INF * INF;
const int MAX_N = 1e3 + 10;
int N, Nr[MAX_N][5];
int main() {
  cin >> N;
  for (int i = 1; i <= N; i++)
    for (int j = 0; j < 5; j++) scanf("%d", &Nr[i][j]);
  if (N >= 100) {
    puts("0");
    return 0;
  }
  vector<int> Ans;
  for (int i = 1; i <= N; i++) {
    bool isCan = true;
    for (int j = 1; j <= N; j++)
      if (j != i)
        for (int k = j + 1; k <= N; k++)
          if (k != i) {
            int sum = 0;
            for (int b = 0; b < 5; b++)
              sum += (Nr[j][b] - Nr[i][b]) * (Nr[k][b] - Nr[i][b]);
            if (sum > 0) {
              isCan = false;
              goto finish;
            }
          }
  finish:
    if (isCan) Ans.push_back(i);
  }
  printf("%d\n", ((int)(Ans).size()));
  for (int x : Ans) printf("%d\n", x);
  return 0;
}

```