#include <bits/stdc++.h>
using namespace std;
int a[1000][5], n;
int b[1000][1000];
int x, y;
int q = 1;
int c = 1000000000;
int main() {
  cin >> n;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 5; j++) cin >> a[i][j];
  for (int i = 1; i < n; i++) {
    b[i][i] = 0;
    for (int j = 0; j < i; j++) {
      b[i][j] = 0;
      for (int k = 0; k < 5; k++) {
        b[i][j] += pow(a[i][k] - a[j][k], 2);
      }
      b[j][i] = b[i][j];
      if (c > b[i][j]) {
        c = b[i][j];
        x = i;
        y = j;
      }
    }
  }
  if (n == 1)
    cout << 1 << "\n" << 1;
  else {
    if (n == 2)
      cout << 2 << "\n" << 1 << "\n" << 2;
    else {
      for (int i = 0; i < n; i++) {
        if (i != x && i != y) {
          if (b[x][y] > b[y][i] - b[x][i]) {
            x = y;
            break;
          }
        }
      }
      for (int i = 1; i < n; i++) {
        if (i != x) {
          for (int j = 0; j < i; j++) {
            if (j != x) {
              if (b[x][i] + b[x][j] > b[i][j]) {
                cout << 0;
                q = 0;
                break;
              }
            }
          }
          if (q == 0) break;
        }
      }
      if (q == 1) cout << 1 << "\n" << x + 1;
    }
  }
}

### Prompt

Please formulate a cpp solution to the following problem:

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 a[1000][5], n;
int b[1000][1000];
int x, y;
int q = 1;
int c = 1000000000;
int main() {
  cin >> n;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 5; j++) cin >> a[i][j];
  for (int i = 1; i < n; i++) {
    b[i][i] = 0;
    for (int j = 0; j < i; j++) {
      b[i][j] = 0;
      for (int k = 0; k < 5; k++) {
        b[i][j] += pow(a[i][k] - a[j][k], 2);
      }
      b[j][i] = b[i][j];
      if (c > b[i][j]) {
        c = b[i][j];
        x = i;
        y = j;
      }
    }
  }
  if (n == 1)
    cout << 1 << "\n" << 1;
  else {
    if (n == 2)
      cout << 2 << "\n" << 1 << "\n" << 2;
    else {
      for (int i = 0; i < n; i++) {
        if (i != x && i != y) {
          if (b[x][y] > b[y][i] - b[x][i]) {
            x = y;
            break;
          }
        }
      }
      for (int i = 1; i < n; i++) {
        if (i != x) {
          for (int j = 0; j < i; j++) {
            if (j != x) {
              if (b[x][i] + b[x][j] > b[i][j]) {
                cout << 0;
                q = 0;
                break;
              }
            }
          }
          if (q == 0) break;
        }
      }
      if (q == 1) cout << 1 << "\n" << x + 1;
    }
  }
}

```