#include <bits/stdc++.h>
using namespace std;
int X[1005][5];
int main() {
  int N;
  scanf("%d", &N);
  for (int i = 0; i < N; ++i) {
    for (int j = 0; j < 5; ++j) scanf("%d", &X[i][j]);
  }
  vector<int> ans;
  for (int i = 0; i < N; ++i) {
    bool bad = false;
    if (N > 100) {
      bad = true;
      goto end;
    }
    for (int j = 0; j < N; ++j)
      if (i != j)
        for (int k = j + 1; k < N; ++k)
          if (i != k) {
            int dot = 0;
            for (int l = 0; l < 5; ++l)
              dot += (X[j][l] - X[i][l]) * (X[k][l] - X[i][l]);
            if (dot > 0) {
              bad = true;
              goto end;
            }
          }
  end:;
    if (!bad) ans.push_back(i + 1);
  }
  printf("%d\n", ans.size());
  for (int a : ans) printf("%d\n", a);
}

### Prompt

Your task is to create 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 X[1005][5];
int main() {
  int N;
  scanf("%d", &N);
  for (int i = 0; i < N; ++i) {
    for (int j = 0; j < 5; ++j) scanf("%d", &X[i][j]);
  }
  vector<int> ans;
  for (int i = 0; i < N; ++i) {
    bool bad = false;
    if (N > 100) {
      bad = true;
      goto end;
    }
    for (int j = 0; j < N; ++j)
      if (i != j)
        for (int k = j + 1; k < N; ++k)
          if (i != k) {
            int dot = 0;
            for (int l = 0; l < 5; ++l)
              dot += (X[j][l] - X[i][l]) * (X[k][l] - X[i][l]);
            if (dot > 0) {
              bad = true;
              goto end;
            }
          }
  end:;
    if (!bad) ans.push_back(i + 1);
  }
  printf("%d\n", ans.size());
  for (int a : ans) printf("%d\n", a);
}

```