#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
int A[N][10];
bool Acute[N];
int n;
double sqr(int k) { return k * k; }
double dis(int u, int v) {
  return sqrt(sqr(A[u][1] - A[v][1]) + sqr(A[u][2] - A[v][2]) +
              sqr(A[u][3] - A[v][3]) + sqr(A[u][v] - A[v][4]) +
              sqr(A[u][5] - A[v][5]));
}
bool check(int r, int u, int v) {
  double ans = 0;
  for (int i = 1; i <= 5; i++) ans += (A[u][i] - A[r][i]) * (A[v][i] - A[r][i]);
  ans /= (dis(r, u) * dis(v, u));
  return acos(ans) * 180 / 3.14159265 < 90;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    for (int k = 1; k <= 5; k++) cin >> A[i][k];
  }
  for (int i = 1; i <= n; i++)
    if (!Acute[i]) {
      for (int j = 1; j <= n; j++) {
        for (int k = j + 1; k <= n; k++)
          if (j != i && k != i) {
            if (check(i, j, k)) {
              Acute[i] = true;
              break;
            } else {
              Acute[j] = Acute[k] = true;
            }
          }
        if (Acute[i]) break;
      }
    }
  int ans = 0;
  for (int i = 1; i <= n; i++) ans += !Acute[i];
  cout << ans << endl;
  for (int i = 1; i <= n; i++)
    if (!Acute[i]) cout << i << endl;
}

### 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;
const int N = 1010;
int A[N][10];
bool Acute[N];
int n;
double sqr(int k) { return k * k; }
double dis(int u, int v) {
  return sqrt(sqr(A[u][1] - A[v][1]) + sqr(A[u][2] - A[v][2]) +
              sqr(A[u][3] - A[v][3]) + sqr(A[u][v] - A[v][4]) +
              sqr(A[u][5] - A[v][5]));
}
bool check(int r, int u, int v) {
  double ans = 0;
  for (int i = 1; i <= 5; i++) ans += (A[u][i] - A[r][i]) * (A[v][i] - A[r][i]);
  ans /= (dis(r, u) * dis(v, u));
  return acos(ans) * 180 / 3.14159265 < 90;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    for (int k = 1; k <= 5; k++) cin >> A[i][k];
  }
  for (int i = 1; i <= n; i++)
    if (!Acute[i]) {
      for (int j = 1; j <= n; j++) {
        for (int k = j + 1; k <= n; k++)
          if (j != i && k != i) {
            if (check(i, j, k)) {
              Acute[i] = true;
              break;
            } else {
              Acute[j] = Acute[k] = true;
            }
          }
        if (Acute[i]) break;
      }
    }
  int ans = 0;
  for (int i = 1; i <= n; i++) ans += !Acute[i];
  cout << ans << endl;
  for (int i = 1; i <= n; i++)
    if (!Acute[i]) cout << i << endl;
}

```