#include <bits/stdc++.h>
using namespace std;
const int N = 1005;
struct point {
  int a, b, c, d, e;
  void got() { scanf("%d%d%d%d%d", &a, &b, &c, &d, &e); }
  bool operator!=(point x) {
    return a != x.a || b != x.b || c != x.c || d != x.d || e != x.e;
  }
} a[N];
struct vec {
  int a, b, c, d, e;
} z;
vec operator-(point a, point b) {
  return (vec){b.a - a.a, b.b - a.b, b.c - a.c, b.d - a.d, b.e - a.e};
}
int operator*(vec a, vec b) {
  return a.a * b.a + a.b * b.b + a.c * b.c + a.d * b.d + a.e * b.e;
}
int n, i, j, k, ans[N], xb;
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) a[i].got();
  if (n > 50) return puts("0"), 0;
  for (i = 1; i <= n; ++i) {
    for (j = 1; j <= n; ++j)
      if (a[i] != a[j]) {
        z = a[j] - a[i];
        for (k = 1; k <= n; ++k)
          if (a[j] != a[k] && a[k] != a[i] && z * (a[k] - a[i]) > 0) break;
        if (k <= n) break;
      }
    if (j > n) ans[++xb] = i;
  }
  printf("%d\n", xb);
  for (i = 1; i <= xb; ++i) printf("%d\n", ans[i]);
  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 int N = 1005;
struct point {
  int a, b, c, d, e;
  void got() { scanf("%d%d%d%d%d", &a, &b, &c, &d, &e); }
  bool operator!=(point x) {
    return a != x.a || b != x.b || c != x.c || d != x.d || e != x.e;
  }
} a[N];
struct vec {
  int a, b, c, d, e;
} z;
vec operator-(point a, point b) {
  return (vec){b.a - a.a, b.b - a.b, b.c - a.c, b.d - a.d, b.e - a.e};
}
int operator*(vec a, vec b) {
  return a.a * b.a + a.b * b.b + a.c * b.c + a.d * b.d + a.e * b.e;
}
int n, i, j, k, ans[N], xb;
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) a[i].got();
  if (n > 50) return puts("0"), 0;
  for (i = 1; i <= n; ++i) {
    for (j = 1; j <= n; ++j)
      if (a[i] != a[j]) {
        z = a[j] - a[i];
        for (k = 1; k <= n; ++k)
          if (a[j] != a[k] && a[k] != a[i] && z * (a[k] - a[i]) > 0) break;
        if (k <= n) break;
      }
    if (j > n) ans[++xb] = i;
  }
  printf("%d\n", xb);
  for (i = 1; i <= xb; ++i) printf("%d\n", ans[i]);
  return 0;
}

```