#include<bits/stdc++.h>
#define REP(i,n) for(int i=0,i##_len=int(n);i<i##_len;++i)
#define rep(i,a,b) for(int i=int(a);i<int(b);++i)
#define All(x) (x).begin(),(x).end()
#define rAll(x) (x).rbegin(),(x).rend()
using namespace std;
using ll = long long;

int main(){
    ll N,K;cin>>N>>K;
    vector<ll> A(N);
    REP(i, N) cin >> A[i];
    sort(All(A));
    ll l=-2e18,r=2e18;
    while(l+1<r){
        ll x=(l+r)>>1;
        ll cnt=0;
        REP(i,N){
            int ok=i,ng=N;
            if(A[i]<0) swap(ok,ng);
            while(abs(ok-ng)>1){
                int midi=(ok+ng)>>1;
                if(A[i]*A[midi]<=x) ok=midi;
                else ng=midi;
            }
            if(A[i]>=0) cnt+=ok-i;
            else cnt+=N-ok;
        }
        if(cnt>=K) r=x;
        else l=x;
    }
    cout<<r<<endl;
}

### 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<bits/stdc++.h>
#define REP(i,n) for(int i=0,i##_len=int(n);i<i##_len;++i)
#define rep(i,a,b) for(int i=int(a);i<int(b);++i)
#define All(x) (x).begin(),(x).end()
#define rAll(x) (x).rbegin(),(x).rend()
using namespace std;
using ll = long long;

int main(){
    ll N,K;cin>>N>>K;
    vector<ll> A(N);
    REP(i, N) cin >> A[i];
    sort(All(A));
    ll l=-2e18,r=2e18;
    while(l+1<r){
        ll x=(l+r)>>1;
        ll cnt=0;
        REP(i,N){
            int ok=i,ng=N;
            if(A[i]<0) swap(ok,ng);
            while(abs(ok-ng)>1){
                int midi=(ok+ng)>>1;
                if(A[i]*A[midi]<=x) ok=midi;
                else ng=midi;
            }
            if(A[i]>=0) cnt+=ok-i;
            else cnt+=N-ok;
        }
        if(cnt>=K) r=x;
        else l=x;
    }
    cout<<r<<endl;
}

```