#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int N = 200000 + 5;
int to;
int n;
ll k;
int a[N];

ll cmp(ll val) {
	int p=to+1;
	ll cnt=0;
	for(int i=to;i>=1 && p<=n;i--) {
		while(val<0 && p<=n && abs((ll)a[i]*a[p])<abs(val)) p++;
		cnt+=max(0,n-p+1);
	}
	return cnt;
}

ll in(int l,int r,ll val) {
	int p=r;
	ll cnt=0;
	for(int i=l;i<p;i++) {
		while(p>0 && (ll)a[i]*a[p]>val) p--;
		cnt+=max(0,p-i);
	}
	return cnt;
}

ll cnt(ll val) {
	return in(1,to,val)+in(to+1,n,val)+cmp(val);
}

int main() {
	cin >> n >> k;
	for(int i = 1; i <= n; i++)
		cin >> a[i];
	sort(a+1,a+n+1);
	for(to = 1; to <= n; to++)
		if(a[to] >= 0)
			break;
	reverse(a+1,a+to);
	to--;
	ll l = -(ll)inf*inf,r = -l;
	while(l <= r){
		ll mid = (l+r)>>1;
		if(cnt(mid) >= k)
			r = mid-1;
		else
			l = mid+1;
	}
	cout << l << endl;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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 inf = 0x3f3f3f3f;
const int N = 200000 + 5;
int to;
int n;
ll k;
int a[N];

ll cmp(ll val) {
	int p=to+1;
	ll cnt=0;
	for(int i=to;i>=1 && p<=n;i--) {
		while(val<0 && p<=n && abs((ll)a[i]*a[p])<abs(val)) p++;
		cnt+=max(0,n-p+1);
	}
	return cnt;
}

ll in(int l,int r,ll val) {
	int p=r;
	ll cnt=0;
	for(int i=l;i<p;i++) {
		while(p>0 && (ll)a[i]*a[p]>val) p--;
		cnt+=max(0,p-i);
	}
	return cnt;
}

ll cnt(ll val) {
	return in(1,to,val)+in(to+1,n,val)+cmp(val);
}

int main() {
	cin >> n >> k;
	for(int i = 1; i <= n; i++)
		cin >> a[i];
	sort(a+1,a+n+1);
	for(to = 1; to <= n; to++)
		if(a[to] >= 0)
			break;
	reverse(a+1,a+to);
	to--;
	ll l = -(ll)inf*inf,r = -l;
	while(l <= r){
		ll mid = (l+r)>>1;
		if(cnt(mid) >= k)
			r = mid-1;
		else
			l = mid+1;
	}
	cout << l << endl;
}
```