#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-9;
struct Point {
  double a, b, c, d, e;
  Point(double a = 0, double b = 0, double c = 0, double d = 0, double e = 0)
      : a(a), b(b), c(c), d(d), e(e) {}
  void read() {
    int v, w, x, y, z;
    scanf("%d%d%d%d%d", &v, &w, &x, &y, &z);
    a = v;
    b = w;
    c = x;
    d = y;
    e = z;
  }
};
inline Point operator-(Point &A, Point &B) {
  return Point(A.a - B.a, A.b - B.b, A.c - B.c, A.d - B.d, A.e - B.e);
}
inline int dcmp(double x) {
  if (fabs(x) < EPS) return 0;
  return (x > 0) ? 1 : -1;
}
inline double Dot(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;
}
inline double len(Point &A) { return sqrt(Dot(A, A)); }
const int MAXN = 1010;
vector<int> ans;
Point p[MAXN];
bool ok[MAXN];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) p[i].read();
  for (int i = 1; i <= n; i++) {
    bool flag = 1;
    for (int j = 1; j <= n && flag; j++) {
      for (int k = 1; k <= n && flag; k++) {
        if (i != j && j != k && i != k) {
          Point l1 = p[i] - p[j];
          Point l2 = p[i] - p[k];
          double len1 = len(l1);
          double len2 = len(l2);
          if (fabs(len1) != 0.0 && fabs(len2) != 0.0) {
            double arc = (Dot(l1, l2) / len1 / len2);
            if (dcmp(arc) > 0) flag = 0;
          }
        }
      }
    }
    if (flag) ans.push_back(i);
  }
  cout << ans.size() << endl;
  for (size_t i = 0; i < ans.size(); i++) cout << ans.at(i) << ' ';
  putchar('\n');
  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 double EPS = 1e-9;
struct Point {
  double a, b, c, d, e;
  Point(double a = 0, double b = 0, double c = 0, double d = 0, double e = 0)
      : a(a), b(b), c(c), d(d), e(e) {}
  void read() {
    int v, w, x, y, z;
    scanf("%d%d%d%d%d", &v, &w, &x, &y, &z);
    a = v;
    b = w;
    c = x;
    d = y;
    e = z;
  }
};
inline Point operator-(Point &A, Point &B) {
  return Point(A.a - B.a, A.b - B.b, A.c - B.c, A.d - B.d, A.e - B.e);
}
inline int dcmp(double x) {
  if (fabs(x) < EPS) return 0;
  return (x > 0) ? 1 : -1;
}
inline double Dot(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;
}
inline double len(Point &A) { return sqrt(Dot(A, A)); }
const int MAXN = 1010;
vector<int> ans;
Point p[MAXN];
bool ok[MAXN];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) p[i].read();
  for (int i = 1; i <= n; i++) {
    bool flag = 1;
    for (int j = 1; j <= n && flag; j++) {
      for (int k = 1; k <= n && flag; k++) {
        if (i != j && j != k && i != k) {
          Point l1 = p[i] - p[j];
          Point l2 = p[i] - p[k];
          double len1 = len(l1);
          double len2 = len(l2);
          if (fabs(len1) != 0.0 && fabs(len2) != 0.0) {
            double arc = (Dot(l1, l2) / len1 / len2);
            if (dcmp(arc) > 0) flag = 0;
          }
        }
      }
    }
    if (flag) ans.push_back(i);
  }
  cout << ans.size() << endl;
  for (size_t i = 0; i < ans.size(); i++) cout << ans.at(i) << ' ';
  putchar('\n');
  return 0;
}

```