#include <bits/stdc++.h>
using namespace std;
const double PI = acos(0.0);
struct node {
  int a, b, c, d, e;
} dis[1009];
int cun[1009];
int n;
bool pan(int k) {
  for (int i = 1; i <= n; i++) {
    if (i != k)
      for (int j = 1; j <= n; j++) {
        if (j != i && j != k) {
          node q, w;
          q.a = dis[i].a - dis[k].a;
          q.b = dis[i].b - dis[k].b;
          q.c = dis[i].c - dis[k].c;
          q.d = dis[i].d - dis[k].d;
          q.e = dis[i].e - dis[k].e;
          w.a = dis[j].a - dis[k].a;
          w.b = dis[j].b - dis[k].b;
          w.c = dis[j].c - dis[k].c;
          w.d = dis[j].d - dis[k].d;
          w.e = dis[j].e - dis[k].e;
          int xy = q.a * w.a + q.b * w.b + q.c * w.c + q.d * w.d + q.e * w.e;
          double x =
              sqrt(q.a * q.a + q.b * q.b + q.c * q.c + q.d * q.d + q.e * q.e);
          double y =
              sqrt(w.a * w.a + w.b * w.b + w.c * w.c + w.d * w.d + w.e * w.e);
          double p = acos((double)xy * 1.0 / x / y);
          if (p < PI) {
            return 0;
          }
        }
      }
  }
  return 1;
}
int main() {
  memset(cun, 0, sizeof(cun));
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> dis[i].a >> dis[i].b >> dis[i].c >> dis[i].d >> dis[i].e;
  }
  int sum = 0;
  for (int i = 1; i <= n; i++) {
    if (pan(i)) {
      sum++;
      cun[sum] = i;
    }
  }
  cout << sum << endl;
  for (int i = 1; i <= sum; i++) cout << cun[i] << endl;
  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;
const double PI = acos(0.0);
struct node {
  int a, b, c, d, e;
} dis[1009];
int cun[1009];
int n;
bool pan(int k) {
  for (int i = 1; i <= n; i++) {
    if (i != k)
      for (int j = 1; j <= n; j++) {
        if (j != i && j != k) {
          node q, w;
          q.a = dis[i].a - dis[k].a;
          q.b = dis[i].b - dis[k].b;
          q.c = dis[i].c - dis[k].c;
          q.d = dis[i].d - dis[k].d;
          q.e = dis[i].e - dis[k].e;
          w.a = dis[j].a - dis[k].a;
          w.b = dis[j].b - dis[k].b;
          w.c = dis[j].c - dis[k].c;
          w.d = dis[j].d - dis[k].d;
          w.e = dis[j].e - dis[k].e;
          int xy = q.a * w.a + q.b * w.b + q.c * w.c + q.d * w.d + q.e * w.e;
          double x =
              sqrt(q.a * q.a + q.b * q.b + q.c * q.c + q.d * q.d + q.e * q.e);
          double y =
              sqrt(w.a * w.a + w.b * w.b + w.c * w.c + w.d * w.d + w.e * w.e);
          double p = acos((double)xy * 1.0 / x / y);
          if (p < PI) {
            return 0;
          }
        }
      }
  }
  return 1;
}
int main() {
  memset(cun, 0, sizeof(cun));
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> dis[i].a >> dis[i].b >> dis[i].c >> dis[i].d >> dis[i].e;
  }
  int sum = 0;
  for (int i = 1; i <= n; i++) {
    if (pan(i)) {
      sum++;
      cun[sum] = i;
    }
  }
  cout << sum << endl;
  for (int i = 1; i <= sum; i++) cout << cun[i] << endl;
  return 0;
}

```