#include <bits/stdc++.h>
using namespace std;
double PI = 3.14159265358979f;
class vec {
 public:
  long long arr[5];
  long long dot(vec next) {
    long long res = 0;
    for (int i = 0; i < 5; i++) res += arr[i] * next.arr[i];
    return res;
  }
  vec operator-(vec next) {
    vec res;
    for (int i = 0; i < 5; i++) res.arr[i] = arr[i] - next.arr[i];
    return res;
  }
};
int main() {
  int n;
  scanf("%d", &n);
  vector<vec> cont;
  for (int i = 0; i < n; i++) {
    vec now;
    for (int j = 0; j < 5; j++) scanf("%lld", &now.arr[j]);
    cont.push_back(now);
  }
  vector<int> ans;
  for (int i = 0; i < n; i++) {
    bool end = false;
    for (int j = 0; !end && j < n; j++) {
      if (i == j) continue;
      for (int k = 0; !end && k < n; k++) {
        if (i == k || j == k) continue;
        if ((cont[j] - cont[i]).dot(cont[k] - cont[i]) > 0) {
          end = true;
          break;
        }
      }
    }
    if (!end) ans.push_back(i + 1);
  }
  printf("%d\n", ans.size());
  for (auto i : ans) printf("%d\n", i);
}

### 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>
using namespace std;
double PI = 3.14159265358979f;
class vec {
 public:
  long long arr[5];
  long long dot(vec next) {
    long long res = 0;
    for (int i = 0; i < 5; i++) res += arr[i] * next.arr[i];
    return res;
  }
  vec operator-(vec next) {
    vec res;
    for (int i = 0; i < 5; i++) res.arr[i] = arr[i] - next.arr[i];
    return res;
  }
};
int main() {
  int n;
  scanf("%d", &n);
  vector<vec> cont;
  for (int i = 0; i < n; i++) {
    vec now;
    for (int j = 0; j < 5; j++) scanf("%lld", &now.arr[j]);
    cont.push_back(now);
  }
  vector<int> ans;
  for (int i = 0; i < n; i++) {
    bool end = false;
    for (int j = 0; !end && j < n; j++) {
      if (i == j) continue;
      for (int k = 0; !end && k < n; k++) {
        if (i == k || j == k) continue;
        if ((cont[j] - cont[i]).dot(cont[k] - cont[i]) > 0) {
          end = true;
          break;
        }
      }
    }
    if (!end) ans.push_back(i + 1);
  }
  printf("%d\n", ans.size());
  for (auto i : ans) printf("%d\n", i);
}

```