#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1);
struct P {
  int a, b, c, d, e;
  P(int a = 0, int b = 0, int c = 0, int d = 0, int e = 0)
      : a(a), b(b), c(c), d(d), e(e) {}
  void in() { scanf("%d%d%d%d%d", &a, &b, &c, &d, &e); }
} p[2000];
int n;
vector<int> ans;
P operator-(P a, P b) {
  return P(a.a - b.a, a.b - b.b, a.c - b.c, a.d - b.d, a.e - b.e);
}
double operator*(P a, P b) {
  return 1.0 * (a.a * b.a + a.b * b.b + a.c * b.c + a.d * b.d + a.e * b.e);
}
double L(P a) { return sqrt(a * a); }
bool check(P a, P b) { return a * b > 0; }
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) p[i].in();
  for (int i = 1; i <= n; i++) {
    int ok = 0;
    for (int j = 1; j <= n && !ok; j++)
      if (i != j)
        for (int k = 1; k <= n && !ok; k++)
          if (i != k && j != k)
            if (check(p[i] - p[j], p[i] - p[k])) {
              ok = 1;
              break;
            }
    if (!ok) ans.push_back(i);
  }
  cout << ans.size() << "\n";
  for (auto v : ans) printf("%d\n", v);
}

### 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 pi = acos(-1);
struct P {
  int a, b, c, d, e;
  P(int a = 0, int b = 0, int c = 0, int d = 0, int e = 0)
      : a(a), b(b), c(c), d(d), e(e) {}
  void in() { scanf("%d%d%d%d%d", &a, &b, &c, &d, &e); }
} p[2000];
int n;
vector<int> ans;
P operator-(P a, P b) {
  return P(a.a - b.a, a.b - b.b, a.c - b.c, a.d - b.d, a.e - b.e);
}
double operator*(P a, P b) {
  return 1.0 * (a.a * b.a + a.b * b.b + a.c * b.c + a.d * b.d + a.e * b.e);
}
double L(P a) { return sqrt(a * a); }
bool check(P a, P b) { return a * b > 0; }
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) p[i].in();
  for (int i = 1; i <= n; i++) {
    int ok = 0;
    for (int j = 1; j <= n && !ok; j++)
      if (i != j)
        for (int k = 1; k <= n && !ok; k++)
          if (i != k && j != k)
            if (check(p[i] - p[j], p[i] - p[k])) {
              ok = 1;
              break;
            }
    if (!ok) ans.push_back(i);
  }
  cout << ans.size() << "\n";
  for (auto v : ans) printf("%d\n", v);
}

```