#include <bits/stdc++.h>
using namespace std;
long long int A[1007][5];
vector<long long int> ans;
long long int dotProd(long long int i, long long int j, long long int k) {
  long long int l;
  long long int temp = 0;
  for (l = 0; l < 5; l++) {
    temp += (A[j][l] - A[i][l]) * (A[k][l] - A[i][l]);
  }
  return temp;
}
int main() {
  long long int i, j, k, l, m, n, x, y, z, a, b, r;
  scanf("%lld", &n);
  for (i = 0; i < n; i++) {
    scanf("%lld", &A[i][0]);
    scanf("%lld", &A[i][1]);
    scanf("%lld", &A[i][2]);
    scanf("%lld", &A[i][3]);
    scanf("%lld", &A[i][4]);
  }
  if (n > 100) {
    printf("0\n");
    return 0;
  }
  for (i = 0; i < n; i++) {
    bool isgood = true;
    for (j = 0; j < n; j++) {
      if (i == j) continue;
      for (k = 0; k < n; k++) {
        if (i == k || k == j) continue;
        if (dotProd(i, j, k) > 0) isgood = false;
      }
    }
    if (isgood) ans.push_back(i + 1);
  }
  printf("%d\n", (int)ans.size());
  for (i = 0; i < (int)ans.size(); i++) {
    printf("%lld ", ans[i]);
  }
  if ((int)ans.size() != 0) printf("\n");
  return 0;
}

### Prompt

Your task is to create 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;
long long int A[1007][5];
vector<long long int> ans;
long long int dotProd(long long int i, long long int j, long long int k) {
  long long int l;
  long long int temp = 0;
  for (l = 0; l < 5; l++) {
    temp += (A[j][l] - A[i][l]) * (A[k][l] - A[i][l]);
  }
  return temp;
}
int main() {
  long long int i, j, k, l, m, n, x, y, z, a, b, r;
  scanf("%lld", &n);
  for (i = 0; i < n; i++) {
    scanf("%lld", &A[i][0]);
    scanf("%lld", &A[i][1]);
    scanf("%lld", &A[i][2]);
    scanf("%lld", &A[i][3]);
    scanf("%lld", &A[i][4]);
  }
  if (n > 100) {
    printf("0\n");
    return 0;
  }
  for (i = 0; i < n; i++) {
    bool isgood = true;
    for (j = 0; j < n; j++) {
      if (i == j) continue;
      for (k = 0; k < n; k++) {
        if (i == k || k == j) continue;
        if (dotProd(i, j, k) > 0) isgood = false;
      }
    }
    if (isgood) ans.push_back(i + 1);
  }
  printf("%d\n", (int)ans.size());
  for (i = 0; i < (int)ans.size(); i++) {
    printf("%lld ", ans[i]);
  }
  if ((int)ans.size() != 0) printf("\n");
  return 0;
}

```