#include <bits/stdc++.h>
using namespace std;
const int MAX = 11234;
struct Node {
  int v[6];
} store[MAX];
bool operator<(const struct Node &a, const struct Node &b) { a.v[0] < b.v[0]; }
Node operator-(const struct Node &a, const struct Node &b) {
  Node rst;
  for (int i = 0; i < 5; i++) {
    rst.v[i] = a.v[i] - b.v[i];
  }
  return rst;
}
double operator*(const struct Node &a, const struct Node &b) {
  double rst = 0;
  for (int i = 0; i < 5; i++) {
    rst += a.v[i] * b.v[i];
  }
  return rst;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      scanf("%d", &store[i].v[j]);
    }
  }
  vector<Node> data;
  vector<int> ans;
  for (int i = 0; i < n; i++) {
    bool flag = true;
    for (int j = 0; j < n && flag; j++) {
      for (int k = 0; k < n && flag; k++) {
        if (i == j || j == k || i == k) continue;
        Node t1 = store[i] - store[j];
        Node t2 = store[i] - store[k];
        double xx = t1 * t1;
        double yy = t2 * t2;
        double t = acos(t1 * t2 / xx / yy);
        if (t < acos(-1) / 2) flag = false;
      }
    }
    if (flag) ans.push_back(i + 1);
  }
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); i++) {
    printf("%d ", ans[i]);
  }
}

### Prompt

Please provide a CPP coded solution 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 MAX = 11234;
struct Node {
  int v[6];
} store[MAX];
bool operator<(const struct Node &a, const struct Node &b) { a.v[0] < b.v[0]; }
Node operator-(const struct Node &a, const struct Node &b) {
  Node rst;
  for (int i = 0; i < 5; i++) {
    rst.v[i] = a.v[i] - b.v[i];
  }
  return rst;
}
double operator*(const struct Node &a, const struct Node &b) {
  double rst = 0;
  for (int i = 0; i < 5; i++) {
    rst += a.v[i] * b.v[i];
  }
  return rst;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      scanf("%d", &store[i].v[j]);
    }
  }
  vector<Node> data;
  vector<int> ans;
  for (int i = 0; i < n; i++) {
    bool flag = true;
    for (int j = 0; j < n && flag; j++) {
      for (int k = 0; k < n && flag; k++) {
        if (i == j || j == k || i == k) continue;
        Node t1 = store[i] - store[j];
        Node t2 = store[i] - store[k];
        double xx = t1 * t1;
        double yy = t2 * t2;
        double t = acos(t1 * t2 / xx / yy);
        if (t < acos(-1) / 2) flag = false;
      }
    }
    if (flag) ans.push_back(i + 1);
  }
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); i++) {
    printf("%d ", ans[i]);
  }
}

```