#include <bits/stdc++.h>
using namespace std;
int p[1010][10], a[110], P[110][110];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= 5; j++) {
      scanf("%d", &p[i][j]);
    }
  }
  if (n > 11) {
    printf("0\n");
  } else {
    int xx = 0;
    for (int i = 1; i <= n; i++) {
      memset(P, 0, sizeof(P));
      int flag = 1;
      for (int j = 1; j <= n; j++) {
        if (i != j) {
          for (int k = 1; k <= 5; k++) {
            P[j][k] = p[j][k] - p[i][k];
          }
          for (int k = 1; k < j; k++) {
            if (k != i) {
              int qq = 0;
              for (int q = 1; q <= 5; q++) qq += P[k][q] * P[j][q];
              if (qq <= 0)
                continue;
              else {
                flag = 0;
                break;
              }
            }
          }
          if (!flag) break;
        }
      }
      if (flag) a[xx++] = i;
    }
    printf("%d\n", xx);
    for (int i = 0; i < xx; i++) printf("%d\n", a[i]);
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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 p[1010][10], a[110], P[110][110];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= 5; j++) {
      scanf("%d", &p[i][j]);
    }
  }
  if (n > 11) {
    printf("0\n");
  } else {
    int xx = 0;
    for (int i = 1; i <= n; i++) {
      memset(P, 0, sizeof(P));
      int flag = 1;
      for (int j = 1; j <= n; j++) {
        if (i != j) {
          for (int k = 1; k <= 5; k++) {
            P[j][k] = p[j][k] - p[i][k];
          }
          for (int k = 1; k < j; k++) {
            if (k != i) {
              int qq = 0;
              for (int q = 1; q <= 5; q++) qq += P[k][q] * P[j][q];
              if (qq <= 0)
                continue;
              else {
                flag = 0;
                break;
              }
            }
          }
          if (!flag) break;
        }
      }
      if (flag) a[xx++] = i;
    }
    printf("%d\n", xx);
    for (int i = 0; i < xx; i++) printf("%d\n", a[i]);
  }
  return 0;
}

```