#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 50;
struct Point {
  double a, b, c, d, e;
  Point() {}
  Point(double _a, double _b, double _c, double _d, double _e) {
    a = _a;
    b = _b;
    c = _c;
    d = _d;
    e = _e;
  }
  Point operator-(const Point &p) const {
    return Point(p.a - a, p.b - b, p.c - c, p.d - d, p.e - e);
  }
} p[maxn];
double cal(Point a, Point b) {
  return a.a * b.a + a.b * b.b + a.c * b.c + a.d * b.d + a.e * b.e;
}
int main() {
  std::ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> ans;
  for (int i = 1; i <= n; i++) {
    cin >> p[i].a >> p[i].b >> p[i].c >> p[i].d >> p[i].e;
  }
  for (int i = 1; i <= n; i++) {
    bool flag = true;
    for (int j = 1; j <= n; j++) {
      if (!flag) break;
      if (i == j) continue;
      for (int k = 1; k <= n; k++) {
        if (i == k || j == k) continue;
        Point v1 = p[j] - p[i], v2 = p[k] - p[i];
        if (cal(v1, v2) > 0) {
          flag = false;
          break;
        }
      }
    }
    if (flag) ans.push_back(i);
  }
  cout << ans.size() << endl;
  for (auto i : ans) cout << i << endl;
  return 0;
}

### Prompt

In CPP, your task is to solve 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 int maxn = 1e5 + 50;
struct Point {
  double a, b, c, d, e;
  Point() {}
  Point(double _a, double _b, double _c, double _d, double _e) {
    a = _a;
    b = _b;
    c = _c;
    d = _d;
    e = _e;
  }
  Point operator-(const Point &p) const {
    return Point(p.a - a, p.b - b, p.c - c, p.d - d, p.e - e);
  }
} p[maxn];
double cal(Point a, Point b) {
  return a.a * b.a + a.b * b.b + a.c * b.c + a.d * b.d + a.e * b.e;
}
int main() {
  std::ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> ans;
  for (int i = 1; i <= n; i++) {
    cin >> p[i].a >> p[i].b >> p[i].c >> p[i].d >> p[i].e;
  }
  for (int i = 1; i <= n; i++) {
    bool flag = true;
    for (int j = 1; j <= n; j++) {
      if (!flag) break;
      if (i == j) continue;
      for (int k = 1; k <= n; k++) {
        if (i == k || j == k) continue;
        Point v1 = p[j] - p[i], v2 = p[k] - p[i];
        if (cal(v1, v2) > 0) {
          flag = false;
          break;
        }
      }
    }
    if (flag) ans.push_back(i);
  }
  cout << ans.size() << endl;
  for (auto i : ans) cout << i << endl;
  return 0;
}

```