#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-3;
const int SZ = 4000050, INF = 0x7FFFFFFF;
const double PI = acos(-1);
struct nd {
  double a, b, c, d, e;
  nd(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) {}
  nd operator-(nd &rbs) {
    return nd(a - rbs.a, b - rbs.b, c - rbs.c, d - rbs.d, e - rbs.e);
  }
};
double len(nd &x) {
  double res = sqrt(x.a * x.a + x.b * x.b + x.c * x.c + x.d * x.d + x.e * x.e);
  return res;
}
double Dot(nd &x, nd &y) {
  return (x.a * y.a + x.b * y.b + x.c * y.c + x.d * y.d + x.e * y.e) / len(x) /
         len(y);
}
bool test(nd &x, nd &y) {
  double ang = acos(Dot(x, y));
  return ang < (PI / 2) && fabs(ang - (PI / 2)) > EPS;
}
int main() {
  std::ios::sync_with_stdio(0);
  int pnum;
  cin >> pnum;
  vector<nd> vct;
  if (pnum > 11) {
    cout << 0 << endl;
    return 0;
  }
  for (int i = 0; i < pnum; ++i) {
    nd tmp;
    cin >> tmp.a >> tmp.b >> tmp.c >> tmp.d >> tmp.e;
    vct.push_back(tmp);
  }
  vector<int> res;
  for (int i = 0; i < pnum; ++i) {
    bool ok = 1;
    for (int j = 0; j < pnum; ++j) {
      if (j == i) continue;
      for (int k = 0; k < pnum; ++k) {
        if (k == j) continue;
        nd ac = vct[j] - vct[i];
        nd bc = vct[k] - vct[i];
        if (test(ac, bc)) ok = 0;
      }
    }
    if (ok) {
      res.push_back(i + 1);
    }
  }
  cout << res.size() << endl;
  for (int i = 0; i < res.size(); ++i) {
    cout << res[i] << endl;
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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-3;
const int SZ = 4000050, INF = 0x7FFFFFFF;
const double PI = acos(-1);
struct nd {
  double a, b, c, d, e;
  nd(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) {}
  nd operator-(nd &rbs) {
    return nd(a - rbs.a, b - rbs.b, c - rbs.c, d - rbs.d, e - rbs.e);
  }
};
double len(nd &x) {
  double res = sqrt(x.a * x.a + x.b * x.b + x.c * x.c + x.d * x.d + x.e * x.e);
  return res;
}
double Dot(nd &x, nd &y) {
  return (x.a * y.a + x.b * y.b + x.c * y.c + x.d * y.d + x.e * y.e) / len(x) /
         len(y);
}
bool test(nd &x, nd &y) {
  double ang = acos(Dot(x, y));
  return ang < (PI / 2) && fabs(ang - (PI / 2)) > EPS;
}
int main() {
  std::ios::sync_with_stdio(0);
  int pnum;
  cin >> pnum;
  vector<nd> vct;
  if (pnum > 11) {
    cout << 0 << endl;
    return 0;
  }
  for (int i = 0; i < pnum; ++i) {
    nd tmp;
    cin >> tmp.a >> tmp.b >> tmp.c >> tmp.d >> tmp.e;
    vct.push_back(tmp);
  }
  vector<int> res;
  for (int i = 0; i < pnum; ++i) {
    bool ok = 1;
    for (int j = 0; j < pnum; ++j) {
      if (j == i) continue;
      for (int k = 0; k < pnum; ++k) {
        if (k == j) continue;
        nd ac = vct[j] - vct[i];
        nd bc = vct[k] - vct[i];
        if (test(ac, bc)) ok = 0;
      }
    }
    if (ok) {
      res.push_back(i + 1);
    }
  }
  cout << res.size() << endl;
  for (int i = 0; i < res.size(); ++i) {
    cout << res[i] << endl;
  }
  return 0;
}

```