#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
const long long INF = std::numeric_limits<long long>::max();
const int MX = 100001;
const long double PI =
    3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058223172535940812848111745028410270;
int N;
vector<vector<int> > pts;
int dot(const vector<int> &a, const vector<int> &b) {
  int ret = 0;
  for (int i = 0; i < 5; ++i) ret += a[i] * b[i];
  return ret;
}
vector<int> operator-(const vector<int> &a, const vector<int> &b) {
  vector<int> ret(5, 0);
  for (int i = 0; i < 5; ++i) {
    ret[i] = a[i] - b[i];
  }
  return ret;
}
double sz(const vector<int> &a) {
  int ret = 0;
  for (int i = 0; i < 5; ++i) {
    ret += a[i] * a[i];
  }
  return sqrt(ret);
}
long double angle(const vector<int> &a, const vector<int> &b) {
  return acos(((double)dot(a, b) / sz(a)) / sz(b));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N;
  for (int i = 0; i < N; ++i) {
    vector<int> c(5);
    for (int j = 0; j < 5; ++j) {
      cin >> c[j];
    }
    pts.push_back(c);
  }
  vector<int> ats;
  vector<bool> blogas(N, false);
  for (int i = 0; i < N; ++i) {
    for (int j = 0; j < N and !blogas[i]; ++j) {
      if (i == j) continue;
      for (int k = j + 1; k < N and !blogas[i]; ++k) {
        if (i == k or j == k) continue;
        long double ang = angle(pts[j] - pts[i], pts[k] - pts[i]);
        if (ang < PI / 2.0) blogas[i] = true;
      }
    }
    if (!blogas[i]) ats.push_back(i + 1);
  }
  printf("%d\n", ats.size());
  for (auto u : ats) printf("%d\n", u);
}

### 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 MOD = 1000000007;
const long long INF = std::numeric_limits<long long>::max();
const int MX = 100001;
const long double PI =
    3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058223172535940812848111745028410270;
int N;
vector<vector<int> > pts;
int dot(const vector<int> &a, const vector<int> &b) {
  int ret = 0;
  for (int i = 0; i < 5; ++i) ret += a[i] * b[i];
  return ret;
}
vector<int> operator-(const vector<int> &a, const vector<int> &b) {
  vector<int> ret(5, 0);
  for (int i = 0; i < 5; ++i) {
    ret[i] = a[i] - b[i];
  }
  return ret;
}
double sz(const vector<int> &a) {
  int ret = 0;
  for (int i = 0; i < 5; ++i) {
    ret += a[i] * a[i];
  }
  return sqrt(ret);
}
long double angle(const vector<int> &a, const vector<int> &b) {
  return acos(((double)dot(a, b) / sz(a)) / sz(b));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N;
  for (int i = 0; i < N; ++i) {
    vector<int> c(5);
    for (int j = 0; j < 5; ++j) {
      cin >> c[j];
    }
    pts.push_back(c);
  }
  vector<int> ats;
  vector<bool> blogas(N, false);
  for (int i = 0; i < N; ++i) {
    for (int j = 0; j < N and !blogas[i]; ++j) {
      if (i == j) continue;
      for (int k = j + 1; k < N and !blogas[i]; ++k) {
        if (i == k or j == k) continue;
        long double ang = angle(pts[j] - pts[i], pts[k] - pts[i]);
        if (ang < PI / 2.0) blogas[i] = true;
      }
    }
    if (!blogas[i]) ats.push_back(i + 1);
  }
  printf("%d\n", ats.size());
  for (auto u : ats) printf("%d\n", u);
}

```