#include <bits/stdc++.h>
using namespace std;
int n;
int a[1010], b[1010], c[1010], d[1010], e[1010];
int sum(int a, int a1, int b, int b1, int c, int c1, int d, int d1, int e,
        int e1) {
  return a * a1 + b * b1 + c * c1 + d * d1 + e * e1;
}
bool check(int i, int j, int k) {
  int tmp =
      sum(a[j] - a[i], a[k] - a[i], b[j] - b[i], b[k] - b[i], c[j] - c[i],
          c[k] - c[i], d[j] - d[i], d[k] - d[i], e[j] - e[i], e[k] - e[i]);
  return tmp > 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i)
    scanf("%d%d%d%d%d", a + i, b + i, c + i, d + i, e + i);
  if (n > 100)
    printf("%d\n", 0);
  else {
    vector<int> ans;
    for (int i = 1; i <= n; ++i) {
      bool flag = 1;
      for (int j = 1; j < n && flag; ++j)
        for (int k = j + 1; k <= n && flag; ++k)
          if (i != j && i != k) {
            if (check(i, j, k)) flag = 0;
          }
      if (flag) ans.push_back(i);
    }
    printf("%d\n", (int)ans.size());
    for (auto v : ans) printf("%d\n", v);
  }
  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;
int n;
int a[1010], b[1010], c[1010], d[1010], e[1010];
int sum(int a, int a1, int b, int b1, int c, int c1, int d, int d1, int e,
        int e1) {
  return a * a1 + b * b1 + c * c1 + d * d1 + e * e1;
}
bool check(int i, int j, int k) {
  int tmp =
      sum(a[j] - a[i], a[k] - a[i], b[j] - b[i], b[k] - b[i], c[j] - c[i],
          c[k] - c[i], d[j] - d[i], d[k] - d[i], e[j] - e[i], e[k] - e[i]);
  return tmp > 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i)
    scanf("%d%d%d%d%d", a + i, b + i, c + i, d + i, e + i);
  if (n > 100)
    printf("%d\n", 0);
  else {
    vector<int> ans;
    for (int i = 1; i <= n; ++i) {
      bool flag = 1;
      for (int j = 1; j < n && flag; ++j)
        for (int k = j + 1; k <= n && flag; ++k)
          if (i != j && i != k) {
            if (check(i, j, k)) flag = 0;
          }
      if (flag) ans.push_back(i);
    }
    printf("%d\n", (int)ans.size());
    for (auto v : ans) printf("%d\n", v);
  }
  return 0;
}

```