#include <bits/stdc++.h>
using namespace std;
struct node {
  int a, b, c, d, e;
} q[10005];
int result(node x, node y, node z) {
  y.a -= x.a, y.b -= x.b, y.c -= x.c, y.e -= x.e, y.d -= x.d;
  z.a -= x.a, z.b -= x.b, z.c -= x.c, z.e -= x.e, z.d -= x.d;
  return z.a * y.a + z.b * y.b + z.c * y.c + z.d * y.d + z.e * y.e;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d%d%d%d", &q[i].a, &q[i].b, &q[i].c, &q[i].d, &q[i].e);
  }
  if (n > 11) {
    puts("0");
    return 0;
  }
  int cnt = 0;
  int ans[10005];
  for (int i = 0; i < n; i++) {
    int flag = 0;
    for (int j = 0; j < n && !flag; j++) {
      if (i == j) continue;
      for (int k = 0; k < n && !flag; k++) {
        if (i == k || k == j) continue;
        if (result(q[i], q[j], q[k]) > 0) {
          flag = 1;
        }
      }
    }
    if (!flag) {
      ans[cnt++] = i + 1;
    }
  }
  printf("%d\n", cnt);
  for (int i = 0; i < cnt; i++) {
    printf("%d ", ans[i]);
  }
  return 0;
}

### Prompt

Your task is to create 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>
using namespace std;
struct node {
  int a, b, c, d, e;
} q[10005];
int result(node x, node y, node z) {
  y.a -= x.a, y.b -= x.b, y.c -= x.c, y.e -= x.e, y.d -= x.d;
  z.a -= x.a, z.b -= x.b, z.c -= x.c, z.e -= x.e, z.d -= x.d;
  return z.a * y.a + z.b * y.b + z.c * y.c + z.d * y.d + z.e * y.e;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d%d%d%d", &q[i].a, &q[i].b, &q[i].c, &q[i].d, &q[i].e);
  }
  if (n > 11) {
    puts("0");
    return 0;
  }
  int cnt = 0;
  int ans[10005];
  for (int i = 0; i < n; i++) {
    int flag = 0;
    for (int j = 0; j < n && !flag; j++) {
      if (i == j) continue;
      for (int k = 0; k < n && !flag; k++) {
        if (i == k || k == j) continue;
        if (result(q[i], q[j], q[k]) > 0) {
          flag = 1;
        }
      }
    }
    if (!flag) {
      ans[cnt++] = i + 1;
    }
  }
  printf("%d\n", cnt);
  for (int i = 0; i < cnt; i++) {
    printf("%d ", ans[i]);
  }
  return 0;
}

```