#include <bits/stdc++.h>
using namespace std;
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){};
  point operator-(point other) {
    return point(a - other.a, b - other.b, c - other.c, d - other.d,
                 e - other.e);
  }
  double operator*(point& other) {
    return a * other.a + b * other.b + c * other.c + d * other.d + e * other.e;
  }
  double mod() {
    double p = a * a + b * b + c * c + d * d + e * e;
    return sqrt(p);
  }
  bool operator==(point& other) {
    return a == other.a && b == other.b && c == other.c && d == other.d &&
           e == other.e;
  }
};
double const pi = acos(-1);
int main() {
  int n;
  cin >> n;
  vector<point> p(n);
  for (int i = 0; i < n; i++) {
    cin >> p[i].a >> p[i].b >> p[i].c >> p[i].d >> p[i].e;
  }
  int cnt = 0;
  vector<int> good(n, 0);
  for (int k = 0; k < n; k++) {
    if (good[k]) {
      continue;
    }
    bool f = 0;
    for (int i = 0; i < n; i++) {
      if (p[i] == p[k]) {
        continue;
      }
      for (int j = 0; j < n; j++) {
        if (p[j] == p[k] || p[i] == p[j]) {
          continue;
        }
        point p_i = p[i] - p[k], p_j = p[j] - p[k];
        if (acos((p_i * p_j) / (p_i.mod() * p_j.mod())) < pi / 2) {
          f = 1;
          break;
        } else {
          good[i] = -1, good[j] = -1;
        }
      }
      if (f) {
        break;
      }
    }
    if (!f) {
      good[k] = 1;
      cnt++;
    }
  }
  cout << cnt << endl;
  for (int i = 0; i < n; i++) {
    if (good[i] == 1) {
      cout << i + 1 << " ";
    }
  }
  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;
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){};
  point operator-(point other) {
    return point(a - other.a, b - other.b, c - other.c, d - other.d,
                 e - other.e);
  }
  double operator*(point& other) {
    return a * other.a + b * other.b + c * other.c + d * other.d + e * other.e;
  }
  double mod() {
    double p = a * a + b * b + c * c + d * d + e * e;
    return sqrt(p);
  }
  bool operator==(point& other) {
    return a == other.a && b == other.b && c == other.c && d == other.d &&
           e == other.e;
  }
};
double const pi = acos(-1);
int main() {
  int n;
  cin >> n;
  vector<point> p(n);
  for (int i = 0; i < n; i++) {
    cin >> p[i].a >> p[i].b >> p[i].c >> p[i].d >> p[i].e;
  }
  int cnt = 0;
  vector<int> good(n, 0);
  for (int k = 0; k < n; k++) {
    if (good[k]) {
      continue;
    }
    bool f = 0;
    for (int i = 0; i < n; i++) {
      if (p[i] == p[k]) {
        continue;
      }
      for (int j = 0; j < n; j++) {
        if (p[j] == p[k] || p[i] == p[j]) {
          continue;
        }
        point p_i = p[i] - p[k], p_j = p[j] - p[k];
        if (acos((p_i * p_j) / (p_i.mod() * p_j.mod())) < pi / 2) {
          f = 1;
          break;
        } else {
          good[i] = -1, good[j] = -1;
        }
      }
      if (f) {
        break;
      }
    }
    if (!f) {
      good[k] = 1;
      cnt++;
    }
  }
  cout << cnt << endl;
  for (int i = 0; i < n; i++) {
    if (good[i] == 1) {
      cout << i + 1 << " ";
    }
  }
  return 0;
}

```