#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  scanf("%d", &n);
  vector<int> ans;
  int vt[1000][5];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      scanf("%d", &vt[i][j]);
    }
  }
  if (n >= 100) {
    printf("0\n");
  } else {
    vector<int> ans;
    for (int i = 0; i < n; i++) {
      int bad = 0;
      for (int j = 0; j < n; j++) {
        for (int k = 0; k < n; k++) {
          if (i != j && j != k && i != k) {
            long long sum = 0;
            for (int l = 0; l < 5; l++) {
              sum += (vt[j][l] - vt[i][l]) * (vt[k][l] - vt[i][l]);
            }
            if (sum > 0) {
              bad++;
            }
          }
        }
      }
      if (!bad) {
        ans.push_back(i + 1);
      }
    }
    printf("%d\n", (int)ans.size());
    for (int i = 0; i < ans.size(); i++) {
      printf("%d\n", ans[i]);
    }
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 main() {
  int n;
  scanf("%d", &n);
  vector<int> ans;
  int vt[1000][5];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      scanf("%d", &vt[i][j]);
    }
  }
  if (n >= 100) {
    printf("0\n");
  } else {
    vector<int> ans;
    for (int i = 0; i < n; i++) {
      int bad = 0;
      for (int j = 0; j < n; j++) {
        for (int k = 0; k < n; k++) {
          if (i != j && j != k && i != k) {
            long long sum = 0;
            for (int l = 0; l < 5; l++) {
              sum += (vt[j][l] - vt[i][l]) * (vt[k][l] - vt[i][l]);
            }
            if (sum > 0) {
              bad++;
            }
          }
        }
      }
      if (!bad) {
        ans.push_back(i + 1);
      }
    }
    printf("%d\n", (int)ans.size());
    for (int i = 0; i < ans.size(); i++) {
      printf("%d\n", ans[i]);
    }
  }
  return 0;
}

```