#include <bits/stdc++.h>
using namespace std;
int a[12][6], ans[12], cnt = 0;
inline int cal(int p, int q, int d) {
  int i, now = 0;
  for (i = 1; i <= 5; i++) {
    int tp = (a[p][i] - a[d][i]) * (a[q][i] - a[d][i]);
    now += tp;
  }
  return now;
}
int main() {
  int i, j, k, n;
  scanf("%d", &n);
  if (n > 11) {
    puts("0");
    return 0;
  }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= 5; j++) {
      scanf("%d", &a[i][j]);
    }
  }
  for (i = 1; i <= n; i++) {
    int tag = 1;
    for (j = 1; j <= n; j++) {
      for (k = 1; k <= n; k++) {
        if (i == j || j == k || i == k) {
          continue;
        }
        if (cal(j, k, i) > 0) {
          tag = 0;
          break;
        }
      }
      if (tag == 0) {
        break;
      }
    }
    if (tag) {
      ans[++cnt] = i;
    }
  }
  printf("%d\n", cnt);
  for (i = 1; i <= cnt; i++) {
    printf("%d\n", ans[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 a[12][6], ans[12], cnt = 0;
inline int cal(int p, int q, int d) {
  int i, now = 0;
  for (i = 1; i <= 5; i++) {
    int tp = (a[p][i] - a[d][i]) * (a[q][i] - a[d][i]);
    now += tp;
  }
  return now;
}
int main() {
  int i, j, k, n;
  scanf("%d", &n);
  if (n > 11) {
    puts("0");
    return 0;
  }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= 5; j++) {
      scanf("%d", &a[i][j]);
    }
  }
  for (i = 1; i <= n; i++) {
    int tag = 1;
    for (j = 1; j <= n; j++) {
      for (k = 1; k <= n; k++) {
        if (i == j || j == k || i == k) {
          continue;
        }
        if (cal(j, k, i) > 0) {
          tag = 0;
          break;
        }
      }
      if (tag == 0) {
        break;
      }
    }
    if (tag) {
      ans[++cnt] = i;
    }
  }
  printf("%d\n", cnt);
  for (i = 1; i <= cnt; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

```