#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<limits.h>
typedef long long LL;
#define rep(i,N) for(int i=0;i<(LL)N;i++)
int compare(const int*a,const int*b){if(*a<*b)return -1;else if(*a==*b)return 0;else return 1;}
inline LL IN(void)
{
  LL x=0;short f=0,c=getchar();
  while(c<48||c>57)f^=c==45,c=getchar();
  while(c>47&&c<58)x=x*10+c-48,c=getchar();
  return f?-x:x;
}
signed main(void)
{
  int N,NG,OK,BASE,minus;
  LL K,L=-pow(16,16)/16,M,R=LLONG_MAX/3,count;
  N=IN(),K=IN();
  int A[N];
  rep(i,N)A[i]=IN();
  qsort(A,N,sizeof(int),(int(*)(const void*,const void*))compare);
  while(R-L>1)
  {
    M=(L+R)/2,count=0;
    rep(i,N)
    {
      NG=i,OK=N,minus=(A[i]<0);
      while(OK-NG>1)
      {
        BASE=(NG+OK)/2;
        if((LL)A[i]*A[BASE]<=M){if(minus)OK=BASE;else NG=BASE;}
        else{if(minus)NG=BASE;else OK=BASE;}
      }
      count+=(minus)?N-OK:NG-i;
    }
    if(count<K)L=M;else R=M;
  }
  return printf("%lld\n",R),0;
}
### Prompt

Your task is to create a Cpp solution to the following problem:

We have N integers A_1, A_2, ..., A_N.

There are \frac{N(N-1)}{2} ways to choose two of them and form a pair. If we compute the product of each of those pairs and sort the results in ascending order, what will be the K-th number in that list?

Constraints

* All values in input are integers.
* 2 \leq N \leq 2 \times 10^5
* 1 \leq K \leq \frac{N(N-1)}{2}
* -10^9 \leq A_i \leq 10^9\ (1 \leq i \leq N)

Input

Input is given from Standard Input in the following format:


N K
A_1 A_2 \dots A_N


Output

Print the answer.

Examples

Input

4 3
3 3 -4 -2


Output

-6


Input

10 40
5 4 3 2 -1 0 0 0 0 0


Output

6


Input

30 413
-170202098 -268409015 537203564 983211703 21608710 -443999067 -937727165 -97596546 -372334013 398994917 -972141167 798607104 -949068442 -959948616 37909651 0 886627544 -20098238 0 -948955241 0 -214720580 277222296 -18897162 834475626 0 -425610555 110117526 663621752 0


Output

448283280358331064

### Response

```cpp
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<limits.h>
typedef long long LL;
#define rep(i,N) for(int i=0;i<(LL)N;i++)
int compare(const int*a,const int*b){if(*a<*b)return -1;else if(*a==*b)return 0;else return 1;}
inline LL IN(void)
{
  LL x=0;short f=0,c=getchar();
  while(c<48||c>57)f^=c==45,c=getchar();
  while(c>47&&c<58)x=x*10+c-48,c=getchar();
  return f?-x:x;
}
signed main(void)
{
  int N,NG,OK,BASE,minus;
  LL K,L=-pow(16,16)/16,M,R=LLONG_MAX/3,count;
  N=IN(),K=IN();
  int A[N];
  rep(i,N)A[i]=IN();
  qsort(A,N,sizeof(int),(int(*)(const void*,const void*))compare);
  while(R-L>1)
  {
    M=(L+R)/2,count=0;
    rep(i,N)
    {
      NG=i,OK=N,minus=(A[i]<0);
      while(OK-NG>1)
      {
        BASE=(NG+OK)/2;
        if((LL)A[i]*A[BASE]<=M){if(minus)OK=BASE;else NG=BASE;}
        else{if(minus)NG=BASE;else OK=BASE;}
      }
      count+=(minus)?N-OK:NG-i;
    }
    if(count<K)L=M;else R=M;
  }
  return printf("%lld\n",R),0;
}
```