#include <bits/stdc++.h>
using namespace std;
struct point {
  double a, b, c, d, e;
  point(double _a, double _b, double _c, double _d, double _e) {
    a = _a;
    b = _b;
    c = _c;
    d = _d;
    e = _e;
  }
};
double get_dist(point a, point b) {
  point c(a.a - b.a, a.b - b.b, a.c - b.c, a.d - b.d, a.e - b.e);
  return c.a * c.a + c.b * c.b + c.c * c.c + c.d * c.d + c.e * c.e;
}
double dist[1000][1000];
vector<point> arr;
vector<int> arr_b;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    double a, b, c, d, e;
    scanf("%lf %lf %lf %lf %lf", &a, &b, &c, &d, &e);
    arr.push_back(point(a, b, c, d, e));
  }
  for (int i = 0; i < arr.size(); ++i) {
    for (int j = i + 1; j < arr.size(); ++j) {
      dist[i][j] = get_dist(arr[i], arr[j]);
      dist[j][i] = dist[i][j];
    }
  }
  for (int i = 0; i < arr.size(); ++i) {
    bool flag = false;
    for (int j = 0; j < arr.size(); ++j) {
      if (j == i) continue;
      if (flag) break;
      for (int k = 0; k < arr.size(); ++k) {
        if (j == k || i == k) continue;
        if (dist[j][k] < dist[i][j] + dist[i][k]) {
          flag = true;
          break;
        }
      }
    }
    if (flag == false) arr_b.push_back(i);
  }
  printf("%d\n", arr_b.size());
  for (int i = 0; i < arr_b.size(); ++i) {
    printf("%d\n", arr_b[i] + 1);
  }
}

### Prompt

Create a solution in cpp for 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;
struct point {
  double a, b, c, d, e;
  point(double _a, double _b, double _c, double _d, double _e) {
    a = _a;
    b = _b;
    c = _c;
    d = _d;
    e = _e;
  }
};
double get_dist(point a, point b) {
  point c(a.a - b.a, a.b - b.b, a.c - b.c, a.d - b.d, a.e - b.e);
  return c.a * c.a + c.b * c.b + c.c * c.c + c.d * c.d + c.e * c.e;
}
double dist[1000][1000];
vector<point> arr;
vector<int> arr_b;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    double a, b, c, d, e;
    scanf("%lf %lf %lf %lf %lf", &a, &b, &c, &d, &e);
    arr.push_back(point(a, b, c, d, e));
  }
  for (int i = 0; i < arr.size(); ++i) {
    for (int j = i + 1; j < arr.size(); ++j) {
      dist[i][j] = get_dist(arr[i], arr[j]);
      dist[j][i] = dist[i][j];
    }
  }
  for (int i = 0; i < arr.size(); ++i) {
    bool flag = false;
    for (int j = 0; j < arr.size(); ++j) {
      if (j == i) continue;
      if (flag) break;
      for (int k = 0; k < arr.size(); ++k) {
        if (j == k || i == k) continue;
        if (dist[j][k] < dist[i][j] + dist[i][k]) {
          flag = true;
          break;
        }
      }
    }
    if (flag == false) arr_b.push_back(i);
  }
  printf("%d\n", arr_b.size());
  for (int i = 0; i < arr_b.size(); ++i) {
    printf("%d\n", arr_b[i] + 1);
  }
}

```