#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 5;
double PI = acos(0.0);
struct Point {
  int a, b, c, d, e;
} f[250];
int main() {
  int n;
  scanf("%d", &n);
  if (n > 500)
    printf("0\n");
  else {
    for (int i = 0; i < n; i++)
      scanf("%d%d%d%d%d", &f[i].a, &f[i].b, &f[i].c, &f[i].d, &f[i].e);
    vector<int> V;
    for (int i = 0; i < n; i++) {
      int ff = 0;
      for (int j = 0; j < n; j++) {
        Point x;
        if (j == i) continue;
        x.a = f[j].a - f[i].a;
        x.b = f[j].b - f[i].b;
        x.c = f[j].c - f[i].c;
        x.d = f[j].d - f[i].d;
        x.e = f[j].e - f[i].e;
        for (int k = 0; k < n; k++) {
          Point y;
          if (j == i || k == j) continue;
          y.a = f[k].a - f[i].a;
          y.b = f[k].b - f[i].b;
          y.c = f[k].c - f[i].c;
          y.d = f[k].d - f[i].d;
          y.e = f[k].e - f[i].e;
          int z = y.a * x.a + y.b * x.b + y.c * x.c + y.d * x.d + y.e * x.e;
          if (z > 0) {
            ff = 1;
            break;
          }
        }
        if (ff) break;
      }
      if (!ff) V.push_back(i + 1);
    }
    printf("%d\n", (int)V.size());
    for (int i = 0; i < (int)V.size(); i++) printf("%d\n", V[i]);
  }
  return 0;
}

### Prompt

Please formulate 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;
const int N = 1e3 + 5;
double PI = acos(0.0);
struct Point {
  int a, b, c, d, e;
} f[250];
int main() {
  int n;
  scanf("%d", &n);
  if (n > 500)
    printf("0\n");
  else {
    for (int i = 0; i < n; i++)
      scanf("%d%d%d%d%d", &f[i].a, &f[i].b, &f[i].c, &f[i].d, &f[i].e);
    vector<int> V;
    for (int i = 0; i < n; i++) {
      int ff = 0;
      for (int j = 0; j < n; j++) {
        Point x;
        if (j == i) continue;
        x.a = f[j].a - f[i].a;
        x.b = f[j].b - f[i].b;
        x.c = f[j].c - f[i].c;
        x.d = f[j].d - f[i].d;
        x.e = f[j].e - f[i].e;
        for (int k = 0; k < n; k++) {
          Point y;
          if (j == i || k == j) continue;
          y.a = f[k].a - f[i].a;
          y.b = f[k].b - f[i].b;
          y.c = f[k].c - f[i].c;
          y.d = f[k].d - f[i].d;
          y.e = f[k].e - f[i].e;
          int z = y.a * x.a + y.b * x.b + y.c * x.c + y.d * x.d + y.e * x.e;
          if (z > 0) {
            ff = 1;
            break;
          }
        }
        if (ff) break;
      }
      if (!ff) V.push_back(i + 1);
    }
    printf("%d\n", (int)V.size());
    for (int i = 0; i < (int)V.size(); i++) printf("%d\n", V[i]);
  }
  return 0;
}

```