#include <bits/stdc++.h>
using namespace std;
int n, m, k;
struct ver {
  int a, b, c, d, e;
};
ver x[1005];
bool was[1005];
int cur[1005], c[1005];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x[i].a >> x[i].b >> x[i].c >> x[i].d >> x[i].e;
  }
  if (n > 11) {
    cout << 0;
    return 0;
  }
  memset(was, 1, 0);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < n; k++) {
        if (i == j || i == k || j == k) continue;
        ver ij, ik;
        ij.a = x[j].a - x[i].a;
        ij.b = x[j].b - x[i].b;
        ij.c = x[j].c - x[i].c;
        ij.d = x[j].d - x[i].d;
        ij.e = x[j].e - x[i].e;
        ik.a = x[k].a - x[i].a;
        ik.b = x[k].b - x[i].b;
        ik.c = x[k].c - x[i].c;
        ik.d = x[k].d - x[i].d;
        ik.e = x[k].e - x[i].e;
        double ab =
            ij.a * ik.a + ij.b * ik.b + ij.c * ik.c + ij.d * ik.d + ij.e * ik.e;
        double a = sqrt(ij.a * ij.a + ij.b * ij.b + ij.c * ij.c + ij.d * ij.d +
                        ij.e * ij.e);
        double b = sqrt(ik.a * ik.a + ik.b * ik.b + ik.c * ik.c + ik.d * ik.d +
                        ik.e * ik.e);
        double cnt = ab / (a * b);
        cur[i + 1]++;
        if (cnt <= 0) c[i + 1]++;
      }
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++)
    if (c[i] == cur[i]) ans++;
  cout << ans << endl;
  for (int i = 1; i <= n; i++)
    if (cur[i] == c[i]) cout << i << " ";
  return 0;
}

### Prompt

Develop a solution in CPP 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;
int n, m, k;
struct ver {
  int a, b, c, d, e;
};
ver x[1005];
bool was[1005];
int cur[1005], c[1005];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x[i].a >> x[i].b >> x[i].c >> x[i].d >> x[i].e;
  }
  if (n > 11) {
    cout << 0;
    return 0;
  }
  memset(was, 1, 0);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < n; k++) {
        if (i == j || i == k || j == k) continue;
        ver ij, ik;
        ij.a = x[j].a - x[i].a;
        ij.b = x[j].b - x[i].b;
        ij.c = x[j].c - x[i].c;
        ij.d = x[j].d - x[i].d;
        ij.e = x[j].e - x[i].e;
        ik.a = x[k].a - x[i].a;
        ik.b = x[k].b - x[i].b;
        ik.c = x[k].c - x[i].c;
        ik.d = x[k].d - x[i].d;
        ik.e = x[k].e - x[i].e;
        double ab =
            ij.a * ik.a + ij.b * ik.b + ij.c * ik.c + ij.d * ik.d + ij.e * ik.e;
        double a = sqrt(ij.a * ij.a + ij.b * ij.b + ij.c * ij.c + ij.d * ij.d +
                        ij.e * ij.e);
        double b = sqrt(ik.a * ik.a + ik.b * ik.b + ik.c * ik.c + ik.d * ik.d +
                        ik.e * ik.e);
        double cnt = ab / (a * b);
        cur[i + 1]++;
        if (cnt <= 0) c[i + 1]++;
      }
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++)
    if (c[i] == cur[i]) ans++;
  cout << ans << endl;
  for (int i = 1; i <= n; i++)
    if (cur[i] == c[i]) cout << i << " ";
  return 0;
}

```