#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1000 + 10;
const int INF = 1e9 + 7;
const double eps = 1e-8;
const double pi = acos(-1.0);
struct point {
  double a[6];
  void input() {
    for (int i = 1; i <= 5; i++) scanf("%lf", &a[i]);
  }
  friend point operator-(const point &a, const point &b) {
    point res;
    for (int i = 1; i <= 5; i++) {
      res.a[i] = a.a[i] - b.a[i];
    }
    return res;
  }
};
double dot(point x, point y) {
  double res = 0;
  for (int i = 1; i <= 5; i++) {
    res += x.a[i] * y.a[i];
  }
  return res;
}
double angle(point x, point y) {
  double tmp = 0;
  tmp = dot(x, y) / sqrt(dot(x, x) * dot(y, y));
  return acos(tmp);
}
int cmp(double x) {
  if (fabs(x) < eps) return 0;
  if (x > 0) return 1;
  return -1;
}
point ps[MAXN];
int pos;
int n;
int tot = 0;
int res[MAXN];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) ps[i].input();
  if (n >= 33) {
    printf("0\n");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    int ok = 0;
    for (int j = 1; j <= n; j++) {
      for (int k = 1; k <= n; k++) {
        if (i == j || i == k || j == k) continue;
        if (cmp(dot(ps[j] - ps[i], ps[k] - ps[i])) > 0) {
          ok = 1;
          break;
        }
      }
      if (ok) break;
    }
    if (ok == 0) {
      res[tot++] = i;
    }
  }
  printf("%d\n", tot);
  for (int i = 0; i < tot; i++) {
    printf("%d\n", res[i]);
  }
}

### Prompt

Generate a CPP solution to the following problem:

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 = 1000 + 10;
const int INF = 1e9 + 7;
const double eps = 1e-8;
const double pi = acos(-1.0);
struct point {
  double a[6];
  void input() {
    for (int i = 1; i <= 5; i++) scanf("%lf", &a[i]);
  }
  friend point operator-(const point &a, const point &b) {
    point res;
    for (int i = 1; i <= 5; i++) {
      res.a[i] = a.a[i] - b.a[i];
    }
    return res;
  }
};
double dot(point x, point y) {
  double res = 0;
  for (int i = 1; i <= 5; i++) {
    res += x.a[i] * y.a[i];
  }
  return res;
}
double angle(point x, point y) {
  double tmp = 0;
  tmp = dot(x, y) / sqrt(dot(x, x) * dot(y, y));
  return acos(tmp);
}
int cmp(double x) {
  if (fabs(x) < eps) return 0;
  if (x > 0) return 1;
  return -1;
}
point ps[MAXN];
int pos;
int n;
int tot = 0;
int res[MAXN];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) ps[i].input();
  if (n >= 33) {
    printf("0\n");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    int ok = 0;
    for (int j = 1; j <= n; j++) {
      for (int k = 1; k <= n; k++) {
        if (i == j || i == k || j == k) continue;
        if (cmp(dot(ps[j] - ps[i], ps[k] - ps[i])) > 0) {
          ok = 1;
          break;
        }
      }
      if (ok) break;
    }
    if (ok == 0) {
      res[tot++] = i;
    }
  }
  printf("%d\n", tot);
  for (int i = 0; i < tot; i++) {
    printf("%d\n", res[i]);
  }
}

```