#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 100;
int ans[N], vis[N], n;
struct node {
  int a, b, c, d, e;
} x[N];
int check(int i, int j, int k) {
  int sum = (x[j].a - x[i].a) * (x[k].a - x[i].a) +
            (x[j].b - x[i].b) * (x[k].b - x[i].b) +
            (x[j].c - x[i].c) * (x[k].c - x[i].c) +
            (x[j].d - x[i].d) * (x[k].d - x[i].d) +
            (x[j].e - x[i].e) * (x[k].e - x[i].e);
  return sum;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d%d%d%d%d", &x[i].a, &x[i].b, &x[i].c, &x[i].d, &x[i].e);
  }
  for (int i = 1; i <= n; ++i) {
    int t = 0;
    if (vis[i]) continue;
    for (int j = 1; j <= n; ++j) {
      for (int k = 1; k <= n; ++k) {
        if (i == j || j == k || i == k) continue;
        if (check(i, j, k) > 0) {
          t = 1;
          break;
        } else {
          vis[j] = 1;
          vis[k] = 1;
        }
      }
      if (t) break;
    }
    if (t) vis[i] = 1;
  }
  int sum = 0;
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) ans[++sum] = i;
  cout << sum << endl;
  for (int i = 1; i <= sum; ++i) cout << ans[i] << endl;
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 N = 1e3 + 100;
int ans[N], vis[N], n;
struct node {
  int a, b, c, d, e;
} x[N];
int check(int i, int j, int k) {
  int sum = (x[j].a - x[i].a) * (x[k].a - x[i].a) +
            (x[j].b - x[i].b) * (x[k].b - x[i].b) +
            (x[j].c - x[i].c) * (x[k].c - x[i].c) +
            (x[j].d - x[i].d) * (x[k].d - x[i].d) +
            (x[j].e - x[i].e) * (x[k].e - x[i].e);
  return sum;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d%d%d%d%d", &x[i].a, &x[i].b, &x[i].c, &x[i].d, &x[i].e);
  }
  for (int i = 1; i <= n; ++i) {
    int t = 0;
    if (vis[i]) continue;
    for (int j = 1; j <= n; ++j) {
      for (int k = 1; k <= n; ++k) {
        if (i == j || j == k || i == k) continue;
        if (check(i, j, k) > 0) {
          t = 1;
          break;
        } else {
          vis[j] = 1;
          vis[k] = 1;
        }
      }
      if (t) break;
    }
    if (t) vis[i] = 1;
  }
  int sum = 0;
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) ans[++sum] = i;
  cout << sum << endl;
  for (int i = 1; i <= sum; ++i) cout << ans[i] << endl;
  return 0;
}

```