#include <bits/stdc++.h>
using namespace std;
const int maxn = 1003;
int n;
struct node {
  int i;
  int x[5];
} a[maxn];
vector<node> act;
int dot(node &x, node &y, node &z) {
  node d1, d2;
  for (int i = 0; i < 5; i++) {
    d1.x[i] = y.x[i] - x.x[i];
    d2.x[i] = z.x[i] - x.x[i];
  }
  int ret = 0;
  for (int i = 0; i < 5; i++) {
    ret += d1.x[i] * d2.x[i];
  }
  return ret;
}
void work() {
  node x = act[0];
  node y = act[1];
  node z = act[2];
  act.clear();
  if (dot(x, y, z) <= 0) {
    act.push_back(x);
  }
  if (dot(y, x, z) <= 0) {
    act.push_back(y);
  }
  if (dot(z, x, y) <= 0) {
    act.push_back(z);
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      cin >> a[i].x[j];
      a[i].i = i + 1;
    }
  }
  if (n <= 2) {
    cout << n << endl;
    for (int i = 0; i < n; i++) cout << a[i].i << endl;
    return 0;
  }
  act.push_back(a[0]);
  act.push_back(a[1]);
  act.push_back(a[2]);
  work();
  int p = 3;
  while (p < n) {
    while (p < n && act.size() < 3) {
      act.push_back(a[p++]);
    }
    work();
  }
  if (act.size() == 2) {
    act.push_back(a[0]);
    work();
  }
  if (act.size() == 0) {
    cout << 0 << endl;
  } else {
    int ans = act[0].i;
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        if (ans - 1 != i && ans - 1 != j) {
          if (dot(a[ans - 1], a[i], a[j]) > 0) {
            cout << 0 << endl;
            return 0;
          }
        }
      }
    }
    cout << 1 << endl;
    cout << ans << 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 int maxn = 1003;
int n;
struct node {
  int i;
  int x[5];
} a[maxn];
vector<node> act;
int dot(node &x, node &y, node &z) {
  node d1, d2;
  for (int i = 0; i < 5; i++) {
    d1.x[i] = y.x[i] - x.x[i];
    d2.x[i] = z.x[i] - x.x[i];
  }
  int ret = 0;
  for (int i = 0; i < 5; i++) {
    ret += d1.x[i] * d2.x[i];
  }
  return ret;
}
void work() {
  node x = act[0];
  node y = act[1];
  node z = act[2];
  act.clear();
  if (dot(x, y, z) <= 0) {
    act.push_back(x);
  }
  if (dot(y, x, z) <= 0) {
    act.push_back(y);
  }
  if (dot(z, x, y) <= 0) {
    act.push_back(z);
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      cin >> a[i].x[j];
      a[i].i = i + 1;
    }
  }
  if (n <= 2) {
    cout << n << endl;
    for (int i = 0; i < n; i++) cout << a[i].i << endl;
    return 0;
  }
  act.push_back(a[0]);
  act.push_back(a[1]);
  act.push_back(a[2]);
  work();
  int p = 3;
  while (p < n) {
    while (p < n && act.size() < 3) {
      act.push_back(a[p++]);
    }
    work();
  }
  if (act.size() == 2) {
    act.push_back(a[0]);
    work();
  }
  if (act.size() == 0) {
    cout << 0 << endl;
  } else {
    int ans = act[0].i;
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        if (ans - 1 != i && ans - 1 != j) {
          if (dot(a[ans - 1], a[i], a[j]) > 0) {
            cout << 0 << endl;
            return 0;
          }
        }
      }
    }
    cout << 1 << endl;
    cout << ans << endl;
  }
  return 0;
}

```