#include <bits/stdc++.h>
using namespace std;
int x[1000][1000];
int y[12];
int g(int a) {
  int b = 0;
  for (int i = 1; i <= a - 2; i++) b += i;
  return b;
}
int f(int a, int b, int c) {
  int d = 0;
  for (int i = 1; i <= 5; i++) d += (x[b][i] - x[a][i]) * (x[c][i] - x[a][i]);
  if (d <= 0)
    return 1;
  else
    return 0;
}
int main() {
  int a, b = 0, c = 0;
  cin >> a;
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= 5; j++) cin >> x[i][j];
  for (int i = 1; i <= a; i++) {
    b = 0;
    int o = 0;
    for (int j = 1; j <= a; j++) {
      for (int k = 1; k <= a; k++) {
        if (j < k && i != j && i != k) {
          if (f(i, j, k) == 1) b++;
          if (f(i, j, k) == 0) {
            o++;
            break;
          }
        }
      }
      if (o != 0) {
        i++;
        break;
      }
    }
    if (c == 11) {
      cout << 0;
      return 0;
    }
    if (b == g(a)) {
      c++;
      y[c] = i;
    }
  }
  cout << c << endl;
  for (int i = 1; i <= c; i++) cout << y[i] << " ";
  return 0;
}

### 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;
int x[1000][1000];
int y[12];
int g(int a) {
  int b = 0;
  for (int i = 1; i <= a - 2; i++) b += i;
  return b;
}
int f(int a, int b, int c) {
  int d = 0;
  for (int i = 1; i <= 5; i++) d += (x[b][i] - x[a][i]) * (x[c][i] - x[a][i]);
  if (d <= 0)
    return 1;
  else
    return 0;
}
int main() {
  int a, b = 0, c = 0;
  cin >> a;
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= 5; j++) cin >> x[i][j];
  for (int i = 1; i <= a; i++) {
    b = 0;
    int o = 0;
    for (int j = 1; j <= a; j++) {
      for (int k = 1; k <= a; k++) {
        if (j < k && i != j && i != k) {
          if (f(i, j, k) == 1) b++;
          if (f(i, j, k) == 0) {
            o++;
            break;
          }
        }
      }
      if (o != 0) {
        i++;
        break;
      }
    }
    if (c == 11) {
      cout << 0;
      return 0;
    }
    if (b == g(a)) {
      c++;
      y[c] = i;
    }
  }
  cout << c << endl;
  for (int i = 1; i <= c; i++) cout << y[i] << " ";
  return 0;
}

```