#include <bits/stdc++.h>
using namespace std;
class vec5 {
 public:
  int a[5];
  void read() {
    for (int i = 0; i < 5; ++i) {
      scanf("%d", a + i);
    }
  }
  int inner(vec5 ii) {
    int res = 0;
    for (int i = 0; i < 5; ++i) {
      res += this->a[i] * ii.a[i];
    }
    return res;
  }
  void minus(vec5 ii, vec5 ip) {
    for (int i = 0; i < 5; ++i) {
      this->a[i] = ii.a[i] - ip.a[i];
    }
  }
};
vec5 arr[260];
int ok[260];
int main() {
  int n;
  scanf("%d", &n);
  if (n > 250) {
    puts("0");
  } else {
    int i;
    for (i = 0; i < n; ++i) {
      arr[i].read();
    }
    int cc = 0;
    for (i = 0; i < n; ++i) {
      int j, k;
      bool lok = true;
      for (j = 0; lok && j < n; ++j) {
        if (j == i) continue;
        for (k = j + 1; k < n; ++k) {
          if (k == i) continue;
          vec5 a, b;
          a.minus(arr[j], arr[i]);
          b.minus(arr[k], arr[i]);
          if (a.inner(b) > 0) {
            lok = false;
            break;
          }
        }
      }
      if (lok) {
        ok[i] = true;
        ++cc;
      }
    }
    printf("%d\n", cc);
    for (i = 0; i < n; ++i) {
      if (ok[i]) {
        printf("%d\n", i + 1);
      }
    }
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
class vec5 {
 public:
  int a[5];
  void read() {
    for (int i = 0; i < 5; ++i) {
      scanf("%d", a + i);
    }
  }
  int inner(vec5 ii) {
    int res = 0;
    for (int i = 0; i < 5; ++i) {
      res += this->a[i] * ii.a[i];
    }
    return res;
  }
  void minus(vec5 ii, vec5 ip) {
    for (int i = 0; i < 5; ++i) {
      this->a[i] = ii.a[i] - ip.a[i];
    }
  }
};
vec5 arr[260];
int ok[260];
int main() {
  int n;
  scanf("%d", &n);
  if (n > 250) {
    puts("0");
  } else {
    int i;
    for (i = 0; i < n; ++i) {
      arr[i].read();
    }
    int cc = 0;
    for (i = 0; i < n; ++i) {
      int j, k;
      bool lok = true;
      for (j = 0; lok && j < n; ++j) {
        if (j == i) continue;
        for (k = j + 1; k < n; ++k) {
          if (k == i) continue;
          vec5 a, b;
          a.minus(arr[j], arr[i]);
          b.minus(arr[k], arr[i]);
          if (a.inner(b) > 0) {
            lok = false;
            break;
          }
        }
      }
      if (lok) {
        ok[i] = true;
        ++cc;
      }
    }
    printf("%d\n", cc);
    for (i = 0; i < n; ++i) {
      if (ok[i]) {
        printf("%d\n", i + 1);
      }
    }
  }
  return 0;
}

```