#include <bits/stdc++.h>
using namespace std;
int n;
int pts[1010][5];
bool bad[1010];
int dot(int x, int y, int z) {
  int prod = 0;
  for (int i = 0; i < 5; i++) {
    prod += (pts[y][i] - pts[x][i]) * (pts[z][i] - pts[x][i]);
  }
  return prod;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      cin >> pts[i][j];
    }
  }
  for (int i = 0; i < n; i++) {
    if (bad[i]) {
      continue;
    }
    bool stop = false;
    for (int j = 0; j < n; j++) {
      if (i == j) {
        continue;
      }
      for (int k = 0; k < n; k++) {
        if (i == k || j == k) {
          continue;
        }
        int prod = dot(i, j, k);
        if (prod > 0) {
          bad[i] = true;
          stop = true;
          break;
        } else {
          bad[j] = bad[k] = true;
        }
      }
      if (stop) {
        break;
      }
    }
  }
  int count = 0;
  for (int i = 0; i < n; i++) {
    if (!bad[i]) {
      count++;
    }
  }
  cout << count << endl;
  for (int i = 0; i < n; i++) {
    if (!bad[i]) {
      cout << i + 1 << endl;
    }
  }
}

### Prompt

Create a solution in cpp for 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 n;
int pts[1010][5];
bool bad[1010];
int dot(int x, int y, int z) {
  int prod = 0;
  for (int i = 0; i < 5; i++) {
    prod += (pts[y][i] - pts[x][i]) * (pts[z][i] - pts[x][i]);
  }
  return prod;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      cin >> pts[i][j];
    }
  }
  for (int i = 0; i < n; i++) {
    if (bad[i]) {
      continue;
    }
    bool stop = false;
    for (int j = 0; j < n; j++) {
      if (i == j) {
        continue;
      }
      for (int k = 0; k < n; k++) {
        if (i == k || j == k) {
          continue;
        }
        int prod = dot(i, j, k);
        if (prod > 0) {
          bad[i] = true;
          stop = true;
          break;
        } else {
          bad[j] = bad[k] = true;
        }
      }
      if (stop) {
        break;
      }
    }
  }
  int count = 0;
  for (int i = 0; i < n; i++) {
    if (!bad[i]) {
      count++;
    }
  }
  cout << count << endl;
  for (int i = 0; i < n; i++) {
    if (!bad[i]) {
      cout << i + 1 << endl;
    }
  }
}

```