#include <bits/stdc++.h>
using namespace std;
const int N = 15005;
const int M = 10921000;
const long long INF = (1LL << 40);
const int MOD = 1e9 + 7;
const double eps = 1e-4;
const double Pi = acos(-1);
int n, m;
struct nd {
  int a, b, c, d, e;
} s[N];
int ans[N];
nd mi(nd x, nd y) {
  return (nd){y.a - x.a, y.b - x.b, y.c - x.c, y.d - x.d, y.e - x.e};
}
int conc(nd x, nd y) {
  return y.a * x.a + y.b * x.b + y.c * x.c + y.d * x.d + y.e * x.e;
}
void pt(nd x) {
  cout << x.a << " " << x.b << " " << x.c << " " << x.d << " " << x.e << " "
       << endl;
}
int main() {
  while (~scanf("%d", &n)) {
    int pos = 0;
    for (int i = 1; i <= n; i++)
      scanf("%d%d%d%d%d", &s[i].a, &s[i].b, &s[i].c, &s[i].d, &s[i].e);
    for (int i = 1; i <= n; i++) {
      int f = 0;
      for (int j = 1; j <= n && f == 0; j++)
        if (j != i)
          for (int k = 1; k <= n && f == 0; k++) {
            if (k == j || k == i) continue;
            nd t1 = mi(s[i], s[j]);
            nd t2 = mi(s[i], s[k]);
            if (acos(conc(t1, t2) / (sqrt(conc(t1, t1)) * sqrt(conc(t2, t2)))) <
                Pi / 2)
              f = 1;
          }
      if (f == 0) ans[pos++] = i;
    }
    printf("%d\n", pos);
    for (int i = 0; i < pos; i++) {
      printf("%d", ans[i]);
      if (i < pos - 1)
        printf(" ");
      else
        printf("\n");
    }
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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 N = 15005;
const int M = 10921000;
const long long INF = (1LL << 40);
const int MOD = 1e9 + 7;
const double eps = 1e-4;
const double Pi = acos(-1);
int n, m;
struct nd {
  int a, b, c, d, e;
} s[N];
int ans[N];
nd mi(nd x, nd y) {
  return (nd){y.a - x.a, y.b - x.b, y.c - x.c, y.d - x.d, y.e - x.e};
}
int conc(nd x, nd y) {
  return y.a * x.a + y.b * x.b + y.c * x.c + y.d * x.d + y.e * x.e;
}
void pt(nd x) {
  cout << x.a << " " << x.b << " " << x.c << " " << x.d << " " << x.e << " "
       << endl;
}
int main() {
  while (~scanf("%d", &n)) {
    int pos = 0;
    for (int i = 1; i <= n; i++)
      scanf("%d%d%d%d%d", &s[i].a, &s[i].b, &s[i].c, &s[i].d, &s[i].e);
    for (int i = 1; i <= n; i++) {
      int f = 0;
      for (int j = 1; j <= n && f == 0; j++)
        if (j != i)
          for (int k = 1; k <= n && f == 0; k++) {
            if (k == j || k == i) continue;
            nd t1 = mi(s[i], s[j]);
            nd t2 = mi(s[i], s[k]);
            if (acos(conc(t1, t2) / (sqrt(conc(t1, t1)) * sqrt(conc(t2, t2)))) <
                Pi / 2)
              f = 1;
          }
      if (f == 0) ans[pos++] = i;
    }
    printf("%d\n", pos);
    for (int i = 0; i < pos; i++) {
      printf("%d", ans[i]);
      if (i < pos - 1)
        printf(" ");
      else
        printf("\n");
    }
  }
  return 0;
}

```