#include <bits/stdc++.h>
using namespace std;
int n;
int ans;
struct P {
  int a[5];
  int f;
} p[1005];
int F(P a, P b, P c) {
  long long sum = 0;
  for (int i = 0; i < 5; i++) {
    sum += (b.a[i] - a.a[i]) * (c.a[i] - a.a[i]);
  }
  if (sum <= 0) return 1;
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 5; j++) {
      scanf("%d", &p[i].a[j]);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (p[i].f == 0) {
      for (int j = 1; j <= n; j++) {
        if (j == i) continue;
        for (int k = 1; k <= n; k++) {
          if (k == i || k == j) continue;
          int ff = F(p[i], p[j], p[k]);
          if (ff) {
            p[j].f = 1;
            p[k].f = 1;
          } else {
            if (p[j].f == 0) {
              if (F(p[j], p[i], p[k])) {
                p[k].f = 1;
              }
            }
            if (p[k].f == 0) {
              if (F(p[k], p[i], p[j])) {
                p[j].f = 1;
              }
            }
            p[i].f = 1;
            break;
          }
        }
        if (p[i].f) break;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (p[i].f == 0) ans++;
  }
  cout << ans << endl;
  int fff = 0;
  for (int i = 1; i <= n; i++) {
    if (p[i].f == 0) {
      if (fff == 0) {
        cout << i;
        fff = 1;
      } else {
        cout << " " << i;
      }
    }
  }
  cout << endl;
  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;
int n;
int ans;
struct P {
  int a[5];
  int f;
} p[1005];
int F(P a, P b, P c) {
  long long sum = 0;
  for (int i = 0; i < 5; i++) {
    sum += (b.a[i] - a.a[i]) * (c.a[i] - a.a[i]);
  }
  if (sum <= 0) return 1;
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 5; j++) {
      scanf("%d", &p[i].a[j]);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (p[i].f == 0) {
      for (int j = 1; j <= n; j++) {
        if (j == i) continue;
        for (int k = 1; k <= n; k++) {
          if (k == i || k == j) continue;
          int ff = F(p[i], p[j], p[k]);
          if (ff) {
            p[j].f = 1;
            p[k].f = 1;
          } else {
            if (p[j].f == 0) {
              if (F(p[j], p[i], p[k])) {
                p[k].f = 1;
              }
            }
            if (p[k].f == 0) {
              if (F(p[k], p[i], p[j])) {
                p[j].f = 1;
              }
            }
            p[i].f = 1;
            break;
          }
        }
        if (p[i].f) break;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (p[i].f == 0) ans++;
  }
  cout << ans << endl;
  int fff = 0;
  for (int i = 1; i <= n; i++) {
    if (p[i].f == 0) {
      if (fff == 0) {
        cout << i;
        fff = 1;
      } else {
        cout << " " << i;
      }
    }
  }
  cout << endl;
  return 0;
}

```