#include <bits/stdc++.h>
using namespace std;
const long double eps = 1e-8;
const long double pi = acos(-1);
const long long inf = 0x3f3f3f3f3f3f3f3f;
const int INF = 0x3f3f3f3f;
const int MAX = 2e5 + 10;
const long long mod = 1e9 + 7;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int sgn(long double k) {
  if (k > eps)
    return 1;
  else if (k < -eps)
    return -1;
  return 0;
}
struct point {
  long double x[5];
};
point sub(point p1, point p2) {
  point ans;
  for (int i = (0); i < (5); ++i) {
    ans.x[i] = p1.x[i] - p2.x[i];
  }
  return ans;
}
long double len(point p) {
  long double ans = 0;
  for (int i = (0); i < (5); ++i) {
    ans += p.x[i] * p.x[i];
  }
  return sqrt(ans);
}
long double cross(point p1, point p2) {
  long double ans = 0;
  for (int i = (0); i < (5); ++i) {
    ans += p1.x[i] * p2.x[i];
  }
  return ans;
}
long double angle(point p1, point p2) {
  return acos(cross(p1, p2) / len(p1) / len(p2));
}
point p[1010];
int sum[1010];
int main() {
  int n;
  cin >> n;
  for (int i = (0); i < (n); ++i) {
    for (int j = (0); j < (5); ++j) {
      cin >> p[i].x[j];
    }
  }
  if (n > 11) {
    cout << '0';
    return 0;
  }
  for (int i = (0); i < (n); ++i) {
    for (int j = (0); j < (n); ++j) {
      for (int k = (0); k < (n); ++k) {
        if (i == k || k == j || i == j) {
          continue;
        }
        point ans1 = sub(p[j], p[i]);
        point ans2 = sub(p[k], p[i]);
        if (sgn(angle(ans1, ans2) - pi / 2) == -1) {
          sum[i]++;
        }
      }
    }
  }
  vector<int> v;
  int ans = 0;
  for (int i = (0); i < (n); ++i) {
    if (sum[i] == 0) {
      ans++;
      v.push_back(i);
    }
  }
  cout << ans << '\n';
  for (auto i = v.begin(); i != v.end(); i++) {
    cout << (*i) + 1 << ' ';
  }
}

### Prompt

Your challenge is to write a Cpp solution to 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 long double eps = 1e-8;
const long double pi = acos(-1);
const long long inf = 0x3f3f3f3f3f3f3f3f;
const int INF = 0x3f3f3f3f;
const int MAX = 2e5 + 10;
const long long mod = 1e9 + 7;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int sgn(long double k) {
  if (k > eps)
    return 1;
  else if (k < -eps)
    return -1;
  return 0;
}
struct point {
  long double x[5];
};
point sub(point p1, point p2) {
  point ans;
  for (int i = (0); i < (5); ++i) {
    ans.x[i] = p1.x[i] - p2.x[i];
  }
  return ans;
}
long double len(point p) {
  long double ans = 0;
  for (int i = (0); i < (5); ++i) {
    ans += p.x[i] * p.x[i];
  }
  return sqrt(ans);
}
long double cross(point p1, point p2) {
  long double ans = 0;
  for (int i = (0); i < (5); ++i) {
    ans += p1.x[i] * p2.x[i];
  }
  return ans;
}
long double angle(point p1, point p2) {
  return acos(cross(p1, p2) / len(p1) / len(p2));
}
point p[1010];
int sum[1010];
int main() {
  int n;
  cin >> n;
  for (int i = (0); i < (n); ++i) {
    for (int j = (0); j < (5); ++j) {
      cin >> p[i].x[j];
    }
  }
  if (n > 11) {
    cout << '0';
    return 0;
  }
  for (int i = (0); i < (n); ++i) {
    for (int j = (0); j < (n); ++j) {
      for (int k = (0); k < (n); ++k) {
        if (i == k || k == j || i == j) {
          continue;
        }
        point ans1 = sub(p[j], p[i]);
        point ans2 = sub(p[k], p[i]);
        if (sgn(angle(ans1, ans2) - pi / 2) == -1) {
          sum[i]++;
        }
      }
    }
  }
  vector<int> v;
  int ans = 0;
  for (int i = (0); i < (n); ++i) {
    if (sum[i] == 0) {
      ans++;
      v.push_back(i);
    }
  }
  cout << ans << '\n';
  for (auto i = v.begin(); i != v.end(); i++) {
    cout << (*i) + 1 << ' ';
  }
}

```