#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e3 + 10;
double n, co[MAXN][5], vec[MAXN][MAXN][5];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) cin >> co[i][j];
  }
  if (n > 11) {
    printf("0\n");
    return 0;
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      for (int first = 0; first < 5; first++) {
        vec[i][j][first] = co[j][first] - co[i][first];
        vec[j][i][first] = co[i][first] - co[j][first];
      }
    }
  }
  vector<int> anses;
  for (int i = 0; i < n; i++) {
    bool can = true;
    for (int a = 0; a < n; a++) {
      if (a == i) continue;
      for (int b = a + 1; b < n; b++) {
        if (b == i) continue;
        int num = 0;
        for (int first = 0; first < 5; first++) {
          num += vec[i][a][first] * vec[i][b][first];
        }
        if (num > 0) can = false;
      }
    }
    if (can) anses.push_back(i + 1);
  }
  printf("%d\n", anses.size());
  for (int v : anses) {
    printf("%d\n", v);
  }
}

### Prompt

Develop a solution in cpp 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;
const int MAXN = 1e3 + 10;
double n, co[MAXN][5], vec[MAXN][MAXN][5];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) cin >> co[i][j];
  }
  if (n > 11) {
    printf("0\n");
    return 0;
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      for (int first = 0; first < 5; first++) {
        vec[i][j][first] = co[j][first] - co[i][first];
        vec[j][i][first] = co[i][first] - co[j][first];
      }
    }
  }
  vector<int> anses;
  for (int i = 0; i < n; i++) {
    bool can = true;
    for (int a = 0; a < n; a++) {
      if (a == i) continue;
      for (int b = a + 1; b < n; b++) {
        if (b == i) continue;
        int num = 0;
        for (int first = 0; first < 5; first++) {
          num += vec[i][a][first] * vec[i][b][first];
        }
        if (num > 0) can = false;
      }
    }
    if (can) anses.push_back(i + 1);
  }
  printf("%d\n", anses.size());
  for (int v : anses) {
    printf("%d\n", v);
  }
}

```