#include <bits/stdc++.h>
using namespace std;
struct Point {
  int x[5];
  Point(int x1 = 0, int x2 = 0, int x3 = 0, int x4 = 0, int x5 = 0) {
    x[0] = x1;
    x[1] = x2;
    x[2] = x3;
    x[3] = x4;
    x[4] = x5;
  }
  Point operator-(const Point& t) {
    return Point(x[0] - t.x[0], x[1] - t.x[1], x[2] - t.x[2], x[3] - t.x[3],
                 x[4] - t.x[4]);
  }
} p[1010];
inline int dot(Point a, Point b) {
  int res = 0;
  for (int i = 0; i < 5; ++i) res += a.x[i] * b.x[i];
  return res;
}
int n;
vector<int> ans;
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i)
    scanf("%d%d%d%d%d", &p[i].x[0], &p[i].x[1], &p[i].x[2], &p[i].x[3],
          &p[i].x[4]);
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j)
      if (j != i) {
        for (int k = 1; k <= n; ++k)
          if (k != j && k != i) {
            if (dot(p[j] - p[i], p[k] - p[i]) > 0) goto HELL;
          }
      }
    ans.push_back(i);
  HELL:;
  }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); ++i) printf("%d\n", ans[i]);
  return 0;
}

### 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 {
  int x[5];
  Point(int x1 = 0, int x2 = 0, int x3 = 0, int x4 = 0, int x5 = 0) {
    x[0] = x1;
    x[1] = x2;
    x[2] = x3;
    x[3] = x4;
    x[4] = x5;
  }
  Point operator-(const Point& t) {
    return Point(x[0] - t.x[0], x[1] - t.x[1], x[2] - t.x[2], x[3] - t.x[3],
                 x[4] - t.x[4]);
  }
} p[1010];
inline int dot(Point a, Point b) {
  int res = 0;
  for (int i = 0; i < 5; ++i) res += a.x[i] * b.x[i];
  return res;
}
int n;
vector<int> ans;
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i)
    scanf("%d%d%d%d%d", &p[i].x[0], &p[i].x[1], &p[i].x[2], &p[i].x[3],
          &p[i].x[4]);
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j)
      if (j != i) {
        for (int k = 1; k <= n; ++k)
          if (k != j && k != i) {
            if (dot(p[j] - p[i], p[k] - p[i]) > 0) goto HELL;
          }
      }
    ans.push_back(i);
  HELL:;
  }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); ++i) printf("%d\n", ans[i]);
  return 0;
}

```