#include <bits/stdc++.h>
using namespace std;
long long n;
long long arr[1005][6];
long long brr[6];
long long crr[6];
vector<long long> v;
int main() {
  scanf("%I64d", &n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      scanf("%I64d", &arr[i][j]);
    }
  }
  long long answer = 0;
  for (int i = 0; i < n; i++) {
    bool can = true;
    for (int j = 0; j < n; j++) {
      if (can == false) break;
      for (int k = j + 1; k < n; k++) {
        if (j == i) break;
        if (k == i) continue;
        for (int l = 0; l < 5; l++) {
          brr[l] = arr[j][l] - arr[i][l];
          crr[l] = arr[k][l] - arr[i][l];
        }
        long long bmod = 0;
        long long cmod = 0;
        long long ddot = 0;
        for (int l = 0; l < 5; l++) {
          bmod += brr[l] * brr[l];
          cmod += crr[l] * crr[l];
          ddot += brr[l] * crr[l];
        }
        if (ddot > 0) {
          can = false;
          break;
        }
      }
    }
    if (can) {
      answer++;
      v.push_back(i + 1);
    }
  }
  printf("%I64d\n", answer);
  for (int i = 0; i < answer; i++) {
    printf("%I64d\n", v[i]);
  }
}

### Prompt

Generate 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;
long long n;
long long arr[1005][6];
long long brr[6];
long long crr[6];
vector<long long> v;
int main() {
  scanf("%I64d", &n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      scanf("%I64d", &arr[i][j]);
    }
  }
  long long answer = 0;
  for (int i = 0; i < n; i++) {
    bool can = true;
    for (int j = 0; j < n; j++) {
      if (can == false) break;
      for (int k = j + 1; k < n; k++) {
        if (j == i) break;
        if (k == i) continue;
        for (int l = 0; l < 5; l++) {
          brr[l] = arr[j][l] - arr[i][l];
          crr[l] = arr[k][l] - arr[i][l];
        }
        long long bmod = 0;
        long long cmod = 0;
        long long ddot = 0;
        for (int l = 0; l < 5; l++) {
          bmod += brr[l] * brr[l];
          cmod += crr[l] * crr[l];
          ddot += brr[l] * crr[l];
        }
        if (ddot > 0) {
          can = false;
          break;
        }
      }
    }
    if (can) {
      answer++;
      v.push_back(i + 1);
    }
  }
  printf("%I64d\n", answer);
  for (int i = 0; i < answer; i++) {
    printf("%I64d\n", v[i]);
  }
}

```