#include <bits/stdc++.h>
using namespace std;
struct node {
  long long int a, b, c, d, e;
} s[11111];
long long int f(long long int a) {
  long long int tp = a * a;
  return tp;
}
int check(node t1, node t2) {
  if (f(t1.a) + f(t1.b) + f(t1.c) + f(t1.d) + f(t1.e) + f(t2.a) + f(t2.b) +
          f(t2.c) + f(t2.d) + f(t2.e) <=
      f(t1.a - t2.a) + f(t1.b - t2.b) + f(t1.c - t2.c) + f(t1.d - t2.d) +
          f(t1.e - t2.e))
    return 1;
  return 0;
}
struct node cx(node x, node y) {
  node t;
  t.a = x.a - y.a;
  t.b = x.b - y.b;
  t.c = x.c - y.c;
  t.d = x.d - y.d;
  t.e = x.e - y.e;
  return t;
}
int main() {
  int n, m, flag, cnt = 0, i, j, k, ji;
  vector<int> tp;
  tp.clear();
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%lld%lld%lld%lld%lld", &s[i].a, &s[i].b, &s[i].c, &s[i].d, &s[i].e);
  }
  for (i = 0; i < n; i++) {
    flag = 1;
    for (j = 0; j < n; j++) {
      if (i == j) continue;
      for (k = 0; k < n; k++) {
        if (k == i || k == j || i == j) continue;
        node tt1 = cx(s[j], s[i]);
        node tt2 = cx(s[k], s[i]);
        if (!check(tt1, tt2)) {
          flag = 0;
          break;
        }
      }
      if (!flag) break;
    }
    if (flag) tp.push_back(i + 1);
  }
  printf("%d\n", tp.size());
  for (i = 0; i < tp.size(); i++) {
    printf("%d ", tp[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;
struct node {
  long long int a, b, c, d, e;
} s[11111];
long long int f(long long int a) {
  long long int tp = a * a;
  return tp;
}
int check(node t1, node t2) {
  if (f(t1.a) + f(t1.b) + f(t1.c) + f(t1.d) + f(t1.e) + f(t2.a) + f(t2.b) +
          f(t2.c) + f(t2.d) + f(t2.e) <=
      f(t1.a - t2.a) + f(t1.b - t2.b) + f(t1.c - t2.c) + f(t1.d - t2.d) +
          f(t1.e - t2.e))
    return 1;
  return 0;
}
struct node cx(node x, node y) {
  node t;
  t.a = x.a - y.a;
  t.b = x.b - y.b;
  t.c = x.c - y.c;
  t.d = x.d - y.d;
  t.e = x.e - y.e;
  return t;
}
int main() {
  int n, m, flag, cnt = 0, i, j, k, ji;
  vector<int> tp;
  tp.clear();
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%lld%lld%lld%lld%lld", &s[i].a, &s[i].b, &s[i].c, &s[i].d, &s[i].e);
  }
  for (i = 0; i < n; i++) {
    flag = 1;
    for (j = 0; j < n; j++) {
      if (i == j) continue;
      for (k = 0; k < n; k++) {
        if (k == i || k == j || i == j) continue;
        node tt1 = cx(s[j], s[i]);
        node tt2 = cx(s[k], s[i]);
        if (!check(tt1, tt2)) {
          flag = 0;
          break;
        }
      }
      if (!flag) break;
    }
    if (flag) tp.push_back(i + 1);
  }
  printf("%d\n", tp.size());
  for (i = 0; i < tp.size(); i++) {
    printf("%d ", tp[i]);
  }
}

```