#include <bits/stdc++.h>
const int N = 1050000;
const int mod = 998244353;
using namespace std;
void io() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
struct Point {
  int a, b, c, d, e;
} p[300007];
bool pp[300007];
long long n, t, m;
int num;
void solve(int i) {
  int a = p[i].a, b = p[i].b, c = p[i].c, d = p[i].d, e = p[i].e;
  for (int j = 1; j <= n; ++j) {
    if (j == i) continue;
    int a1 = a - p[j].a, b1 = b - p[j].b, c1 = c - p[j].c, d1 = d - p[j].d,
        e1 = e - p[j].e;
    for (int k = 1; k <= n; ++k) {
      if ((k == j) || (k == i)) continue;
      int a2 = a - p[k].a, b2 = b - p[k].b, c2 = c - p[k].c, d2 = d - p[k].d,
          e2 = e - p[k].e;
      int tmp = a1 * a2 + b1 * b2 + c1 * c2 + d1 * d2 + e1 * e2;
      if (tmp > 0) {
        pp[i] = true;
        num--;
        break;
      }
    }
    if (pp[i]) break;
  }
}
int main() {
  io();
  cin >> n;
  num = n;
  if (n > 11)
    cout << "0";
  else {
    for (int i = 1; i <= n; ++i) {
      cin >> p[i].a >> p[i].b >> p[i].c >> p[i].d >> p[i].e;
    }
    for (int i = 1; i <= n; ++i) {
      solve(i);
    }
    cout << num << endl;
    for (int i = 1; i <= n; ++i) {
      if (!pp[i]) cout << i << ' ';
    }
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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>
const int N = 1050000;
const int mod = 998244353;
using namespace std;
void io() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
struct Point {
  int a, b, c, d, e;
} p[300007];
bool pp[300007];
long long n, t, m;
int num;
void solve(int i) {
  int a = p[i].a, b = p[i].b, c = p[i].c, d = p[i].d, e = p[i].e;
  for (int j = 1; j <= n; ++j) {
    if (j == i) continue;
    int a1 = a - p[j].a, b1 = b - p[j].b, c1 = c - p[j].c, d1 = d - p[j].d,
        e1 = e - p[j].e;
    for (int k = 1; k <= n; ++k) {
      if ((k == j) || (k == i)) continue;
      int a2 = a - p[k].a, b2 = b - p[k].b, c2 = c - p[k].c, d2 = d - p[k].d,
          e2 = e - p[k].e;
      int tmp = a1 * a2 + b1 * b2 + c1 * c2 + d1 * d2 + e1 * e2;
      if (tmp > 0) {
        pp[i] = true;
        num--;
        break;
      }
    }
    if (pp[i]) break;
  }
}
int main() {
  io();
  cin >> n;
  num = n;
  if (n > 11)
    cout << "0";
  else {
    for (int i = 1; i <= n; ++i) {
      cin >> p[i].a >> p[i].b >> p[i].c >> p[i].d >> p[i].e;
    }
    for (int i = 1; i <= n; ++i) {
      solve(i);
    }
    cout << num << endl;
    for (int i = 1; i <= n; ++i) {
      if (!pp[i]) cout << i << ' ';
    }
  }
  return 0;
}

```