#include <bits/stdc++.h>
const int PI = 3.14159265;
using namespace std;
struct dulieu {
  double a, b, c, d, e;
};
dulieu h[1009], x2, y2;
bool check[1009], checkm;
int kq[1009], res;
void vecx(dulieu y, dulieu z) {
  x2.a = y.a - z.a;
  x2.b = y.b - z.b;
  x2.c = y.c - z.c;
  x2.d = y.d - z.d;
  x2.e = y.e - z.e;
}
void vecy(dulieu y, dulieu z) {
  y2.a = y.a - z.a;
  y2.b = y.b - z.b;
  y2.c = y.c - z.c;
  y2.d = y.d - z.d;
  y2.e = y.e - z.e;
}
int tinh(dulieu x, dulieu y) {
  return (x.a * y.a + x.b * y.b + x.c * y.c + x.d * y.d + x.e * y.e);
}
int tinhl(dulieu x, dulieu y) {
  double d1 = sqrt(x.a * x.a + x.b * x.b + x.c * x.c + x.d * x.d + x.e * x.e);
  double d2 = sqrt(y.a * y.a + y.b * y.b + y.c * y.c + y.d * y.d + y.e * y.e);
  return (d1 * d2);
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++)
    cin >> h[i].a >> h[i].b >> h[i].c >> h[i].d >> h[i].e;
  for (int i = 1; i <= n; i++)
    if (not check[i]) {
      checkm = true;
      for (int j = 1; j <= n; j++) {
        for (int z = 1; z <= n; z++)
          if (i != j && j != z) {
            vecx(h[i], h[j]);
            vecy(h[i], h[z]);
            double res1 = tinh(x2, y2);
            double res2 = tinhl(x2, y2);
            double tam = acos(res1 / res2);
            if (tam * 180.0 / PI < 90) {
              check[j] = true;
              check[z] = true;
              checkm = false;
              break;
            }
          }
        if (not checkm) break;
      }
      if (checkm) {
        ++res;
        kq[res] = i;
      }
    }
  cout << res << endl;
  for (int i = 1; i <= res; i++) cout << kq[i] << endl;
  return 0;
}

### Prompt

Generate a Cpp solution 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>
const int PI = 3.14159265;
using namespace std;
struct dulieu {
  double a, b, c, d, e;
};
dulieu h[1009], x2, y2;
bool check[1009], checkm;
int kq[1009], res;
void vecx(dulieu y, dulieu z) {
  x2.a = y.a - z.a;
  x2.b = y.b - z.b;
  x2.c = y.c - z.c;
  x2.d = y.d - z.d;
  x2.e = y.e - z.e;
}
void vecy(dulieu y, dulieu z) {
  y2.a = y.a - z.a;
  y2.b = y.b - z.b;
  y2.c = y.c - z.c;
  y2.d = y.d - z.d;
  y2.e = y.e - z.e;
}
int tinh(dulieu x, dulieu y) {
  return (x.a * y.a + x.b * y.b + x.c * y.c + x.d * y.d + x.e * y.e);
}
int tinhl(dulieu x, dulieu y) {
  double d1 = sqrt(x.a * x.a + x.b * x.b + x.c * x.c + x.d * x.d + x.e * x.e);
  double d2 = sqrt(y.a * y.a + y.b * y.b + y.c * y.c + y.d * y.d + y.e * y.e);
  return (d1 * d2);
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++)
    cin >> h[i].a >> h[i].b >> h[i].c >> h[i].d >> h[i].e;
  for (int i = 1; i <= n; i++)
    if (not check[i]) {
      checkm = true;
      for (int j = 1; j <= n; j++) {
        for (int z = 1; z <= n; z++)
          if (i != j && j != z) {
            vecx(h[i], h[j]);
            vecy(h[i], h[z]);
            double res1 = tinh(x2, y2);
            double res2 = tinhl(x2, y2);
            double tam = acos(res1 / res2);
            if (tam * 180.0 / PI < 90) {
              check[j] = true;
              check[z] = true;
              checkm = false;
              break;
            }
          }
        if (not checkm) break;
      }
      if (checkm) {
        ++res;
        kq[res] = i;
      }
    }
  cout << res << endl;
  for (int i = 1; i <= res; i++) cout << kq[i] << endl;
  return 0;
}

```