#include <bits/stdc++.h>
using namespace std;
const int N = 200200;
const int mod = 1e9 + 7;
int a[3003][6];
int n;
int e[3003][6];
long long len[3003];
int cur;
int cnt;
int bad[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= 5; ++j) cin >> a[i][j];
  }
  for (int i = 1; i <= n; ++i) {
    if (bad[i]) continue;
    cnt = 0;
    for (int j = 1; j <= n; ++j) {
      len[j] = 0;
      for (int k = 1; k <= 5; ++k) {
        e[j][k] = a[j][k] - a[i][k];
        len[j] += e[j][k] * e[j][k];
      }
    }
    long long f = 0;
    bool ok = 0;
    for (int j = 1; j <= n && !ok; ++j)
      for (int k = j + 1; !ok && k <= n; ++k) {
        if (j == i || k == i) continue;
        f = 0;
        for (int ind = 1; ind <= 5; ++ind) f += e[j][ind] * e[k][ind];
        if (f > 0 && f * f <= len[j] * len[k]) {
          ok = 1;
          bad[i] = 1;
        } else
          bad[j] = bad[k] = 1;
      }
  }
  vector<int> ans;
  for (int i = 1; i <= n; ++i)
    if (!bad[i]) ans.push_back(i);
  cout << ans.size() << "\n";
  for (int x : ans) cout << x << ' ';
  return 0;
}

### 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;
const int N = 200200;
const int mod = 1e9 + 7;
int a[3003][6];
int n;
int e[3003][6];
long long len[3003];
int cur;
int cnt;
int bad[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= 5; ++j) cin >> a[i][j];
  }
  for (int i = 1; i <= n; ++i) {
    if (bad[i]) continue;
    cnt = 0;
    for (int j = 1; j <= n; ++j) {
      len[j] = 0;
      for (int k = 1; k <= 5; ++k) {
        e[j][k] = a[j][k] - a[i][k];
        len[j] += e[j][k] * e[j][k];
      }
    }
    long long f = 0;
    bool ok = 0;
    for (int j = 1; j <= n && !ok; ++j)
      for (int k = j + 1; !ok && k <= n; ++k) {
        if (j == i || k == i) continue;
        f = 0;
        for (int ind = 1; ind <= 5; ++ind) f += e[j][ind] * e[k][ind];
        if (f > 0 && f * f <= len[j] * len[k]) {
          ok = 1;
          bad[i] = 1;
        } else
          bad[j] = bad[k] = 1;
      }
  }
  vector<int> ans;
  for (int i = 1; i <= n; ++i)
    if (!bad[i]) ans.push_back(i);
  cout << ans.size() << "\n";
  for (int x : ans) cout << x << ' ';
  return 0;
}

```