#include <bits/stdc++.h>
using namespace std;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int maxn = 1000 + 10;
const int mod = 1000000007;
struct Point {
  double a, b, c, d, e;
} dian[maxn];
vector<pair<int, Point> > hh[maxn];
vector<int> Ans;
Point cul(Point x1, Point x2) {
  Point ans;
  ans.a = x2.a - x1.a;
  ans.b = x2.b - x1.b;
  ans.c = x2.c - x1.c;
  ans.d = x2.d - x1.d;
  ans.e = x2.e - x1.e;
  return ans;
}
double mult(Point x1, Point x2) {
  return x1.a * x2.a + x1.b * x2.b + x1.c * x2.c + x1.d * x2.d + x1.e * x2.e;
}
bool eque(double x1, double x2) {
  if (fabs(x1 - x2) < 0.0000000001)
    return true;
  else
    return false;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> dian[i].a >> dian[i].b >> dian[i].c >> dian[i].d >> dian[i].e;
  }
  for (int i = 1; i <= n; i++) {
    int flag = 1;
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      for (int k = 1; k <= n; k++) {
        if (k == i || k == j) continue;
        Point ab = cul(dian[i], dian[j]);
        Point ac = cul(dian[i], dian[k]);
        double xy = mult(ab, ac);
        if (xy > 0) {
          flag = 0;
          break;
        }
      }
      if (!flag) {
        break;
      }
    }
    if (flag) {
      Ans.push_back(i);
    }
  }
  cout << Ans.size() << endl;
  for (int i = 0; i < Ans.size(); i++) {
    cout << Ans[i] << endl;
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 long long INF = 0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int maxn = 1000 + 10;
const int mod = 1000000007;
struct Point {
  double a, b, c, d, e;
} dian[maxn];
vector<pair<int, Point> > hh[maxn];
vector<int> Ans;
Point cul(Point x1, Point x2) {
  Point ans;
  ans.a = x2.a - x1.a;
  ans.b = x2.b - x1.b;
  ans.c = x2.c - x1.c;
  ans.d = x2.d - x1.d;
  ans.e = x2.e - x1.e;
  return ans;
}
double mult(Point x1, Point x2) {
  return x1.a * x2.a + x1.b * x2.b + x1.c * x2.c + x1.d * x2.d + x1.e * x2.e;
}
bool eque(double x1, double x2) {
  if (fabs(x1 - x2) < 0.0000000001)
    return true;
  else
    return false;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> dian[i].a >> dian[i].b >> dian[i].c >> dian[i].d >> dian[i].e;
  }
  for (int i = 1; i <= n; i++) {
    int flag = 1;
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      for (int k = 1; k <= n; k++) {
        if (k == i || k == j) continue;
        Point ab = cul(dian[i], dian[j]);
        Point ac = cul(dian[i], dian[k]);
        double xy = mult(ab, ac);
        if (xy > 0) {
          flag = 0;
          break;
        }
      }
      if (!flag) {
        break;
      }
    }
    if (flag) {
      Ans.push_back(i);
    }
  }
  cout << Ans.size() << endl;
  for (int i = 0; i < Ans.size(); i++) {
    cout << Ans[i] << endl;
  }
  return 0;
}

```