#include <bits/stdc++.h>
using namespace std;
const int MM = 1e9 + 7;
const double eps = 1e-8;
const int MAXN = 2e6 + 10;
int n, m;
void prework() {}
void read() {}
long long a[1111][10];
long long dot(int x, int j, int k) {
  long long ret = 0;
  for (int i = 1; i <= 5; i++) ret += (a[j][i] - a[x][i]) * (a[k][i] - a[x][i]);
  return ret;
}
int check(int x) {
  for (int j = 1; j <= n; j++)
    if (j != x)
      for (int k = j + 1; k <= n; k++)
        if (k != x) {
          if (dot(x, j, k) > 0) return 0;
        }
  return 1;
}
void solve(int casi) {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= 5; j++) scanf("%lld", &a[i][j]);
  }
  vector<int> ans;
  for (int i = 1; i <= n; i++) {
    if (check(i)) ans.push_back(i);
  }
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); i++)
    printf("%d%c", ans[i], " \n"[i + 1 == ans.size()]);
}
void printans() {}
int main() {
  prework();
  int T = 1;
  for (int i = 1; i <= T; i++) {
    read();
    solve(i);
    printans();
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 MM = 1e9 + 7;
const double eps = 1e-8;
const int MAXN = 2e6 + 10;
int n, m;
void prework() {}
void read() {}
long long a[1111][10];
long long dot(int x, int j, int k) {
  long long ret = 0;
  for (int i = 1; i <= 5; i++) ret += (a[j][i] - a[x][i]) * (a[k][i] - a[x][i]);
  return ret;
}
int check(int x) {
  for (int j = 1; j <= n; j++)
    if (j != x)
      for (int k = j + 1; k <= n; k++)
        if (k != x) {
          if (dot(x, j, k) > 0) return 0;
        }
  return 1;
}
void solve(int casi) {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= 5; j++) scanf("%lld", &a[i][j]);
  }
  vector<int> ans;
  for (int i = 1; i <= n; i++) {
    if (check(i)) ans.push_back(i);
  }
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); i++)
    printf("%d%c", ans[i], " \n"[i + 1 == ans.size()]);
}
void printans() {}
int main() {
  prework();
  int T = 1;
  for (int i = 1; i <= T; i++) {
    read();
    solve(i);
    printans();
  }
  return 0;
}

```