#include <bits/stdc++.h>
using namespace std;
const double Pi = acos(-1);
const int MAXN = 1e3 + 10;
int n;
struct ee {
  double a, b, c, d, e;
} e[MAXN << 1];
queue<int> tq;
bool isgood(int a, int b, int c) {
  if (a == b) return false;
  if (a == c) return false;
  ee tp;
  double x1 = e[b].a - e[a].a;
  double x2 = e[b].b - e[a].b;
  double x3 = e[b].c - e[a].c;
  double x4 = e[b].d - e[a].d;
  double x5 = e[b].e - e[a].e;
  double y1 = e[c].a - e[a].a;
  double y2 = e[c].b - e[a].b;
  double y3 = e[c].c - e[a].c;
  double y4 = e[c].d - e[a].d;
  double y5 = e[c].e - e[a].e;
  double up = x1 * y1 + x2 * y2 + x3 * y3 + x4 * y4 + x5 * y5;
  double d1 = x1 * x1 + x2 * x2 + x3 * x3 + x4 * x4 + x5 * x5;
  double d2 = y1 * y1 + y2 * y2 + y3 * y3 + y4 * y4 + y5 * y5;
  double down = sqrt(d1 * d2);
  double ang = acos(up / down);
  if (ang < Pi / 2) return true;
  return false;
}
int main() {
  while (cin >> n) {
    for (int i = 1; i <= n; ++i) {
      cin >> e[i].a >> e[i].b >> e[i].c >> e[i].d >> e[i].e;
      tq.push(i);
    }
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= n; ++j) {
        if (i == j) continue;
        int sz = tq.size();
        for (int k = 1; k <= sz; ++k) {
          auto tmp = tq.front();
          tq.pop();
          if (isgood(tmp, i, j)) {
          } else {
            tq.push(tmp);
          }
        }
      }
    }
    cout << tq.size() << endl;
    if (tq.size()) {
      while (tq.size()) {
        cout << tq.front() << " ";
        tq.pop();
      }
      cout << endl;
    }
  }
  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>
using namespace std;
const double Pi = acos(-1);
const int MAXN = 1e3 + 10;
int n;
struct ee {
  double a, b, c, d, e;
} e[MAXN << 1];
queue<int> tq;
bool isgood(int a, int b, int c) {
  if (a == b) return false;
  if (a == c) return false;
  ee tp;
  double x1 = e[b].a - e[a].a;
  double x2 = e[b].b - e[a].b;
  double x3 = e[b].c - e[a].c;
  double x4 = e[b].d - e[a].d;
  double x5 = e[b].e - e[a].e;
  double y1 = e[c].a - e[a].a;
  double y2 = e[c].b - e[a].b;
  double y3 = e[c].c - e[a].c;
  double y4 = e[c].d - e[a].d;
  double y5 = e[c].e - e[a].e;
  double up = x1 * y1 + x2 * y2 + x3 * y3 + x4 * y4 + x5 * y5;
  double d1 = x1 * x1 + x2 * x2 + x3 * x3 + x4 * x4 + x5 * x5;
  double d2 = y1 * y1 + y2 * y2 + y3 * y3 + y4 * y4 + y5 * y5;
  double down = sqrt(d1 * d2);
  double ang = acos(up / down);
  if (ang < Pi / 2) return true;
  return false;
}
int main() {
  while (cin >> n) {
    for (int i = 1; i <= n; ++i) {
      cin >> e[i].a >> e[i].b >> e[i].c >> e[i].d >> e[i].e;
      tq.push(i);
    }
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= n; ++j) {
        if (i == j) continue;
        int sz = tq.size();
        for (int k = 1; k <= sz; ++k) {
          auto tmp = tq.front();
          tq.pop();
          if (isgood(tmp, i, j)) {
          } else {
            tq.push(tmp);
          }
        }
      }
    }
    cout << tq.size() << endl;
    if (tq.size()) {
      while (tq.size()) {
        cout << tq.front() << " ";
        tq.pop();
      }
      cout << endl;
    }
  }
  return 0;
}

```