#include <bits/stdc++.h>
using namespace std;
struct fd {
  int x1, x2, x3, x4, x5;
};
fd tmp, a[1009];
int i, n, j, l;
vector<int> res;
bool ok;
void read(fd &x) { cin >> x.x1 >> x.x2 >> x.x3 >> x.x4 >> x.x5; }
int mu(fd x, fd y) {
  return x.x1 * y.x1 + x.x2 * y.x2 + x.x3 * y.x3 + x.x4 * y.x4 + x.x5 * y.x5;
}
fd mi(fd x, fd y) {
  tmp.x1 = y.x1 - x.x1;
  tmp.x2 = y.x2 - x.x2;
  tmp.x3 = y.x3 - x.x3;
  tmp.x4 = y.x4 - x.x4;
  tmp.x5 = y.x5 - x.x5;
  return tmp;
}
int main() {
  cin >> n;
  for (i = 1; i <= n; i++) {
    read(a[i]);
  }
  for (i = 1; i <= n; i++) {
    ok = true;
    for (j = 1; j <= n; j++) {
      for (l = 1; l <= n; l++) {
        if ((i != j) && (j != l) && (i != l)) {
          if ((acos(mu(mi(a[i], a[j]), mi(a[i], a[l])) /
                    (sqrt(mu(mi(a[i], a[j]), mi(a[i], a[j]))) *
                     sqrt(mu(mi(a[i], a[l]), mi(a[i], a[l]))))) *
               180.0 / 3.14159265) < 90) {
            ok = false;
            break;
          }
        }
      }
      if (!ok) {
        break;
      }
    }
    if (ok) {
      res.push_back(i);
    }
  }
  cout << res.size() << '\n';
  for (i = 0; i < res.size(); i++) {
    cout << res[i] << ' ';
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
struct fd {
  int x1, x2, x3, x4, x5;
};
fd tmp, a[1009];
int i, n, j, l;
vector<int> res;
bool ok;
void read(fd &x) { cin >> x.x1 >> x.x2 >> x.x3 >> x.x4 >> x.x5; }
int mu(fd x, fd y) {
  return x.x1 * y.x1 + x.x2 * y.x2 + x.x3 * y.x3 + x.x4 * y.x4 + x.x5 * y.x5;
}
fd mi(fd x, fd y) {
  tmp.x1 = y.x1 - x.x1;
  tmp.x2 = y.x2 - x.x2;
  tmp.x3 = y.x3 - x.x3;
  tmp.x4 = y.x4 - x.x4;
  tmp.x5 = y.x5 - x.x5;
  return tmp;
}
int main() {
  cin >> n;
  for (i = 1; i <= n; i++) {
    read(a[i]);
  }
  for (i = 1; i <= n; i++) {
    ok = true;
    for (j = 1; j <= n; j++) {
      for (l = 1; l <= n; l++) {
        if ((i != j) && (j != l) && (i != l)) {
          if ((acos(mu(mi(a[i], a[j]), mi(a[i], a[l])) /
                    (sqrt(mu(mi(a[i], a[j]), mi(a[i], a[j]))) *
                     sqrt(mu(mi(a[i], a[l]), mi(a[i], a[l]))))) *
               180.0 / 3.14159265) < 90) {
            ok = false;
            break;
          }
        }
      }
      if (!ok) {
        break;
      }
    }
    if (ok) {
      res.push_back(i);
    }
  }
  cout << res.size() << '\n';
  for (i = 0; i < res.size(); i++) {
    cout << res[i] << ' ';
  }
}

```