#include <bits/stdc++.h>
const int N = 1010;
int n;
int cord[N][5];
bool good[N];
int dot(int sit0, int sit1, int sit2) {
  int ret = 0;
  for (int i = 0; i < 5; ++i) {
    ret += (cord[sit1][i] - cord[sit0][i]) * (cord[sit2][i] - cord[sit0][i]);
  }
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 5; ++j) {
      scanf("%d", &cord[i][j]);
    }
  }
  if (n >= 100) {
    return printf("0\n"), 0;
  }
  int cnt = 0;
  for (int i = 0; i < n; ++i) {
    good[i] = true;
    for (int j = 0; j < n; ++j) {
      if (j == i) continue;
      for (int k = j + 1; k < n; ++k) {
        if (k == i) continue;
        if (dot(i, j, k) > 0) {
          good[i] = false;
          break;
        }
      }
    }
    cnt += good[i];
  }
  printf("%d\n", cnt);
  for (int i = 0; i < n; ++i) {
    if (good[i]) {
      printf("%d\n", i + 1);
    }
  }
  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>
const int N = 1010;
int n;
int cord[N][5];
bool good[N];
int dot(int sit0, int sit1, int sit2) {
  int ret = 0;
  for (int i = 0; i < 5; ++i) {
    ret += (cord[sit1][i] - cord[sit0][i]) * (cord[sit2][i] - cord[sit0][i]);
  }
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 5; ++j) {
      scanf("%d", &cord[i][j]);
    }
  }
  if (n >= 100) {
    return printf("0\n"), 0;
  }
  int cnt = 0;
  for (int i = 0; i < n; ++i) {
    good[i] = true;
    for (int j = 0; j < n; ++j) {
      if (j == i) continue;
      for (int k = j + 1; k < n; ++k) {
        if (k == i) continue;
        if (dot(i, j, k) > 0) {
          good[i] = false;
          break;
        }
      }
    }
    cnt += good[i];
  }
  printf("%d\n", cnt);
  for (int i = 0; i < n; ++i) {
    if (good[i]) {
      printf("%d\n", i + 1);
    }
  }
  return 0;
}

```