#include <bits/stdc++.h>
using namespace std;
const int N = 1001;
int n;
long long v[N][5];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < 5; ++j) scanf("%lld", &v[i][j]);
  if (n > 100)
    puts("0");
  else {
    vector<long long> sol;
    for (int a = 0; a < n; ++a) {
      bool ok = true;
      for (int b = 0; b < n; ++b) {
        for (int c = 0; c < n; ++c) {
          if (a == b || a == c || b == c) continue;
          long long dot = 0;
          for (int k = 0; k < 5; ++k)
            dot += (v[b][k] - v[a][k]) * (v[c][k] - v[a][k]);
          if (dot > 0) ok = false;
        }
      }
      if (ok) sol.push_back(a);
    }
    printf("%d\n", (int)sol.size());
    for (int i = 0; i < (int)sol.size(); ++i) printf("%lld\n", sol[i] + 1);
  }
  return 0;
}

### 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;
const int N = 1001;
int n;
long long v[N][5];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < 5; ++j) scanf("%lld", &v[i][j]);
  if (n > 100)
    puts("0");
  else {
    vector<long long> sol;
    for (int a = 0; a < n; ++a) {
      bool ok = true;
      for (int b = 0; b < n; ++b) {
        for (int c = 0; c < n; ++c) {
          if (a == b || a == c || b == c) continue;
          long long dot = 0;
          for (int k = 0; k < 5; ++k)
            dot += (v[b][k] - v[a][k]) * (v[c][k] - v[a][k]);
          if (dot > 0) ok = false;
        }
      }
      if (ok) sol.push_back(a);
    }
    printf("%d\n", (int)sol.size());
    for (int i = 0; i < (int)sol.size(); ++i) printf("%lld\n", sol[i] + 1);
  }
  return 0;
}

```