#include <bits/stdc++.h>
using namespace std;
int x[1001][5];
bool check(int ni, int nj, int nk) {
  double s12 = 0, s11 = 0, s22 = 0;
  for (int i = 0; i < 5; i++) {
    double s1 = x[nj][i] - x[ni][i];
    double s2 = x[nk][i] - x[ni][i];
    s12 += s1 * s2;
    s11 += s1 * s1;
    s22 += s2 * s2;
  }
  return acos(s12 / sqrt(s11 * s22)) < 0.5 * acos(-1);
}
int main() {
  vector<int> v;
  int n;
  scanf("%d", &n);
  for (int ni = 0; ni < n; ni++)
    for (int i = 0; i < 5; i++) scanf("%d", &x[ni][i]);
  for (int ni = 0; ni < n; ni++) {
    for (int nj = 0; nj < n; nj++)
      for (int nk = 0; nk < n; nk++) {
        if (ni == nj || nj == nk || nk == ni) continue;
        if (check(ni, nj, nk)) goto flag;
      }
    v.push_back(ni + 1);
  flag:;
  }
  printf("%d\n", (int)v.size());
  for (auto num : v) printf("%d\n", num);
}

### Prompt

In Cpp, your task is to solve 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 x[1001][5];
bool check(int ni, int nj, int nk) {
  double s12 = 0, s11 = 0, s22 = 0;
  for (int i = 0; i < 5; i++) {
    double s1 = x[nj][i] - x[ni][i];
    double s2 = x[nk][i] - x[ni][i];
    s12 += s1 * s2;
    s11 += s1 * s1;
    s22 += s2 * s2;
  }
  return acos(s12 / sqrt(s11 * s22)) < 0.5 * acos(-1);
}
int main() {
  vector<int> v;
  int n;
  scanf("%d", &n);
  for (int ni = 0; ni < n; ni++)
    for (int i = 0; i < 5; i++) scanf("%d", &x[ni][i]);
  for (int ni = 0; ni < n; ni++) {
    for (int nj = 0; nj < n; nj++)
      for (int nk = 0; nk < n; nk++) {
        if (ni == nj || nj == nk || nk == ni) continue;
        if (check(ni, nj, nk)) goto flag;
      }
    v.push_back(ni + 1);
  flag:;
  }
  printf("%d\n", (int)v.size());
  for (auto num : v) printf("%d\n", num);
}

```