#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3 + 7;
int dot[maxn][maxn];
int a[maxn], b[maxn], c[maxn], d[maxn], e[maxn];
int len[maxn];
vector<int> good;
int n;
bool flag = false;
bool check(int i) {
  for (int j = 0; j < n; j++)
    for (int k = 0; k < n; k++) {
      if (j == i || k == i || j == k) continue;
      if ((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]) >
          0) {
        return true;
      }
    }
  return false;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++)
    scanf("%d%d%d%d%d", &a[i], &b[i], &c[i], &d[i], &e[i]);
  if (n > 50) {
    printf("0\n");
    return 0;
  }
  for (int i = 0; i < n; i++)
    if (!check(i)) good.push_back(i + 1);
  printf("%d\n", good.size());
  for (int i = 0; i < good.size(); i++) printf("%d\n", good[i]);
}

### Prompt

Generate 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;
const int maxn = 1e3 + 7;
int dot[maxn][maxn];
int a[maxn], b[maxn], c[maxn], d[maxn], e[maxn];
int len[maxn];
vector<int> good;
int n;
bool flag = false;
bool check(int i) {
  for (int j = 0; j < n; j++)
    for (int k = 0; k < n; k++) {
      if (j == i || k == i || j == k) continue;
      if ((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]) >
          0) {
        return true;
      }
    }
  return false;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++)
    scanf("%d%d%d%d%d", &a[i], &b[i], &c[i], &d[i], &e[i]);
  if (n > 50) {
    printf("0\n");
    return 0;
  }
  for (int i = 0; i < n; i++)
    if (!check(i)) good.push_back(i + 1);
  printf("%d\n", good.size());
  for (int i = 0; i < good.size(); i++) printf("%d\n", good[i]);
}

```