#include <bits/stdc++.h>
const int INF = 1e9 + 7;
const long long INFLL = 1e17;
const double EPS = 1e-6;
using namespace std;
const int maxn = 1005;
int n;
int arr[maxn][5];
int main() {
  scanf("%d", &n);
  for (int k = (int)0; k < (int)n; k++)
    for (int i = (int)0; i < (int)5; i++) scanf("%d", &arr[k][i]);
  vector<int> ret;
  for (int k = (int)0; k < (int)n; k++) {
    int bisa = 1;
    for (int i = (int)0; i < (int)n; i++)
      if (i != k) {
        for (int j = (int)i + 1; j < (int)n; j++)
          if (j != k && j != k) {
            int cur = 0;
            for (int a = (int)0; a < (int)5; a++)
              cur += (arr[i][a] - arr[k][a]) * (arr[j][a] - arr[k][a]);
            if (cur > 0) bisa = 0;
            if (!bisa) break;
          }
        if (!bisa) break;
      }
    if (bisa) ret.push_back(k);
  }
  printf("%d\n", ret.size());
  for (auto a : ret) cout << a + 1 << endl;
}

### Prompt

Please provide a cpp coded solution 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>
const int INF = 1e9 + 7;
const long long INFLL = 1e17;
const double EPS = 1e-6;
using namespace std;
const int maxn = 1005;
int n;
int arr[maxn][5];
int main() {
  scanf("%d", &n);
  for (int k = (int)0; k < (int)n; k++)
    for (int i = (int)0; i < (int)5; i++) scanf("%d", &arr[k][i]);
  vector<int> ret;
  for (int k = (int)0; k < (int)n; k++) {
    int bisa = 1;
    for (int i = (int)0; i < (int)n; i++)
      if (i != k) {
        for (int j = (int)i + 1; j < (int)n; j++)
          if (j != k && j != k) {
            int cur = 0;
            for (int a = (int)0; a < (int)5; a++)
              cur += (arr[i][a] - arr[k][a]) * (arr[j][a] - arr[k][a]);
            if (cur > 0) bisa = 0;
            if (!bisa) break;
          }
        if (!bisa) break;
      }
    if (bisa) ret.push_back(k);
  }
  printf("%d\n", ret.size());
  for (auto a : ret) cout << a + 1 << endl;
}

```