#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1);
struct st {
  int a, b, c, d, e;
} a[1005], b[1005][1005];
vector<int> v;
double mul(st x, st y) {
  return (double)(x.a * y.a + x.b * y.b + x.c * y.c + x.d * y.d + x.e * y.e);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++)
    scanf("%d%d%d%d%d", &a[i].a, &a[i].b, &a[i].c, &a[i].d, &a[i].e);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) b[i][j].a = a[i].a - a[j].a;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) b[i][j].b = a[i].b - a[j].b;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) b[i][j].c = a[i].c - a[j].c;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) b[i][j].d = a[i].d - a[j].d;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) b[i][j].e = a[i].e - a[j].e;
  for (int i = 0; i < n; i++) {
    bool f = true;
    for (int j = 0; j < n; j++)
      if (j != i) {
        for (int k = j + 1; k < n; k++) {
          if (k == i) continue;
          int x = mul(b[i][j], b[i][k]);
          if (x > 0) {
            f = false;
            break;
          }
        }
        if (!f) break;
      }
    if (f) v.push_back(i + 1);
  }
  printf("%d\n", v.size());
  for (int i = 0; i < v.size(); i++) printf("%d\n", v[i]);
  return 0;
}

### Prompt

Create a solution in cpp for 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 double pi = acos(-1);
struct st {
  int a, b, c, d, e;
} a[1005], b[1005][1005];
vector<int> v;
double mul(st x, st y) {
  return (double)(x.a * y.a + x.b * y.b + x.c * y.c + x.d * y.d + x.e * y.e);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++)
    scanf("%d%d%d%d%d", &a[i].a, &a[i].b, &a[i].c, &a[i].d, &a[i].e);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) b[i][j].a = a[i].a - a[j].a;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) b[i][j].b = a[i].b - a[j].b;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) b[i][j].c = a[i].c - a[j].c;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) b[i][j].d = a[i].d - a[j].d;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) b[i][j].e = a[i].e - a[j].e;
  for (int i = 0; i < n; i++) {
    bool f = true;
    for (int j = 0; j < n; j++)
      if (j != i) {
        for (int k = j + 1; k < n; k++) {
          if (k == i) continue;
          int x = mul(b[i][j], b[i][k]);
          if (x > 0) {
            f = false;
            break;
          }
        }
        if (!f) break;
      }
    if (f) v.push_back(i + 1);
  }
  printf("%d\n", v.size());
  for (int i = 0; i < v.size(); i++) printf("%d\n", v[i]);
  return 0;
}

```