#include <bits/stdc++.h>
using namespace std;
const long long N = 1e6 + 5;
const long long INF = 1e9 + 7;
struct data {
  long long a, b, c, d, e;
};
long long n;
data a[N];
bool M[N];
vector<long long> V;
inline long long mul(const data &x, const data &y) {
  return x.a * y.a + x.b * y.b + x.c * y.c + x.d * y.d + x.e * y.e;
}
inline data sub(const long long &x, const long long &y) {
  data ret;
  ret.a = a[x].a - a[y].a;
  ret.b = a[x].b - a[y].b;
  ret.c = a[x].c - a[y].c;
  ret.d = a[x].d - a[y].d;
  ret.e = a[x].e - a[y].e;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (long long i = 1; i <= n; i++)
    cin >> a[i].a >> a[i].b >> a[i].c >> a[i].d >> a[i].e;
  memset(M, true, sizeof M);
  for (long long i = 1; i <= n; i++)
    if (M[i]) {
      for (long long j = 1; j <= n && M[i]; j++)
        if (i != j)
          for (long long k = 1; k <= n && M[i]; k++)
            if (k != i && k != j) {
              data fi = sub(j, i), se = sub(k, i);
              long double X = mul(fi, se);
              long double Y = sqrtl(mul(fi, fi));
              long double Z = sqrtl(mul(se, se));
              long double t = X * 1.0 / (Y * Z);
              t = acos(t) * (long double)(180) / (long double)(3.14159265);
              if (t < 90) M[i] = false;
            }
      if (M[i]) V.push_back(i);
    }
  cout << V.size() << endl;
  for (auto it : V) cout << it << endl;
  return 0;
}

### Prompt

Please create a solution in cpp 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 long N = 1e6 + 5;
const long long INF = 1e9 + 7;
struct data {
  long long a, b, c, d, e;
};
long long n;
data a[N];
bool M[N];
vector<long long> V;
inline long long mul(const data &x, const data &y) {
  return x.a * y.a + x.b * y.b + x.c * y.c + x.d * y.d + x.e * y.e;
}
inline data sub(const long long &x, const long long &y) {
  data ret;
  ret.a = a[x].a - a[y].a;
  ret.b = a[x].b - a[y].b;
  ret.c = a[x].c - a[y].c;
  ret.d = a[x].d - a[y].d;
  ret.e = a[x].e - a[y].e;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (long long i = 1; i <= n; i++)
    cin >> a[i].a >> a[i].b >> a[i].c >> a[i].d >> a[i].e;
  memset(M, true, sizeof M);
  for (long long i = 1; i <= n; i++)
    if (M[i]) {
      for (long long j = 1; j <= n && M[i]; j++)
        if (i != j)
          for (long long k = 1; k <= n && M[i]; k++)
            if (k != i && k != j) {
              data fi = sub(j, i), se = sub(k, i);
              long double X = mul(fi, se);
              long double Y = sqrtl(mul(fi, fi));
              long double Z = sqrtl(mul(se, se));
              long double t = X * 1.0 / (Y * Z);
              t = acos(t) * (long double)(180) / (long double)(3.14159265);
              if (t < 90) M[i] = false;
            }
      if (M[i]) V.push_back(i);
    }
  cout << V.size() << endl;
  for (auto it : V) cout << it << endl;
  return 0;
}

```