C. Five Dimensional Points

C. Five Dimensional Points

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 and is acute (i.e. strictly less than ). Otherwise, the point is called good.

The angle between vectors and in 5-dimensional space is defined as , where is the scalar product and is length of .

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 a_i, b_i, c_i, d_i, e_i (|a_i|, |b_i|, |c_i|, |d_i|, |e_i| ≤ 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 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:

We can see that all angles here are acute, so no points are good.

题意：

有n个五维的点，求有多少个“好“点，对于“好”点、“坏”点的定义如下：

“好”点：设该点为a，在所给点中任意两个不相等且不为a的点b,c，向量ab与向量bc的夹角均不为锐角。

“坏”点：不是“好”点的点都是“坏”点

暴力求解;

 

#include <bits/stdc++.h>
#define N 1005
using namespace std;
int cnt=0;
struct Node{
  int a,b,c,d,e;
}k[N];
int as[N];
queue<int> s;
int ssr(Node a,Node b,Node c){ return (a.a-b.a)*(c.a-b.a)+(a.b-b.b)*(c.b-b.b)+(a.c-b.c)*(c.c-b.c)+(a.d-b.d)*(c.d-b.d)+(a.e-b.e)*(c.e-b.e);
}
int main(){
  int n;
  scanf("%d",&n);
  cnt=n;
  for(int i=1;i<=n;i++){
      scanf("%d%d%d%d%d",&k[i].a,&k[i].b,&k[i].c,&k[i].d,&k[i].e);
  }
  memset(as,0,sizeof(as));
  for(int i=1;i<=n;i++){
    for(int j=1;j<=n;j++){
      for(int t=j+1; t<=n;t++){
        if(i!=j&&i!=t&&ssr(k[j],k[i],k[t])>0){
          as[i]=1;
          cnt--;
          goto eg;
        }
      }
    }
    eg:;
  }
  printf("%d
",cnt);
  for(int i=1;i<=n;i++){
    if(as[i]==0)
      printf("%d
",i);
  }
  return 0;
}