#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e5+10;
ll a[N];
ll b[N];
ll tot1,tot2,k;
ll num(ll x){
    ll cnt=0;
    int l=0,r=tot1-1;
    while(l<r){
        while(l<r&&a[l]*a[r]>x)--r;
        cnt+=r-l++;
    }
    l=0,r=tot2-1;
    while(l<r){
        while(l<r&&b[l]*b[r]>x)--r;
        cnt+=r-l++;
    }
    if(x>=0)return cnt+tot1*tot2;
    x=-x;
    l=0;r=tot2-1;
    while(l<tot1){
        while(r>=0&&a[l]*b[r]>=x)--r;
        cnt+=tot2-r-1;
        ++l;
    }
    return cnt;
}
int main(){
    int n;
    cin>>n>>k;
    for(int i=0;i<n;++i){
        ll tmp;
        cin>>tmp;
        if(tmp<0)a[tot1++]=-tmp;
        else b[tot2++]=tmp;
    }
    sort(a,a+tot1);
    sort(b,b+tot2);
    ll l=-1e18-1,r=1e18+1;
    while(l<r){
        ll mid=(l+r)>>1;
        if(num(mid)>=k)r=mid;
        else l=mid+1;
    }
    printf("%lld",l);
    return 0;
}
### Prompt

Please formulate 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>
using namespace std;
typedef long long ll;
const int N=2e5+10;
ll a[N];
ll b[N];
ll tot1,tot2,k;
ll num(ll x){
    ll cnt=0;
    int l=0,r=tot1-1;
    while(l<r){
        while(l<r&&a[l]*a[r]>x)--r;
        cnt+=r-l++;
    }
    l=0,r=tot2-1;
    while(l<r){
        while(l<r&&b[l]*b[r]>x)--r;
        cnt+=r-l++;
    }
    if(x>=0)return cnt+tot1*tot2;
    x=-x;
    l=0;r=tot2-1;
    while(l<tot1){
        while(r>=0&&a[l]*b[r]>=x)--r;
        cnt+=tot2-r-1;
        ++l;
    }
    return cnt;
}
int main(){
    int n;
    cin>>n>>k;
    for(int i=0;i<n;++i){
        ll tmp;
        cin>>tmp;
        if(tmp<0)a[tot1++]=-tmp;
        else b[tot2++]=tmp;
    }
    sort(a,a+tot1);
    sort(b,b+tot2);
    ll l=-1e18-1,r=1e18+1;
    while(l<r){
        ll mid=(l+r)>>1;
        if(num(mid)>=k)r=mid;
        else l=mid+1;
    }
    printf("%lld",l);
    return 0;
}
```