#include <bits/stdc++.h>
using namespace std;

long long n, k;
long long a[200004];

long long f(long long mid){
	long long ret = 0;
	for(int i = 0; i < n; i++){
		if(a[i] == 0){
			if(mid >= 0)	ret += n-1;
			continue;
		}
		if(a[i] < 0){
			long long target = mid/a[i];
			if(target*a[i] > mid)	target++;
			int idx = lower_bound(a, a+n, target)-a;
			ret += n - idx;
			if(idx <= i)	ret -= 1;
		}
		else{
			long long target = mid/a[i];
			if(target*a[i] > mid)	target--;
			int idx = upper_bound(a, a+n, target)-a-1;
			if(idx < 0)	continue;
			ret += idx + 1;
			if(idx >= i)	ret -= 1;
		}
	}
	return ret/2;
}

int main() {
	cin>>n>>k;
	for(int i = 0; i < n; i++)	cin>>a[i];
	sort(a, a+n);
	long long lo = -1e18, hi = 1e18;
	while(lo < hi){
		long long mid = lo + (hi - lo)/2;
		if(f(mid) >= k)	hi = mid;
		else	lo = mid + 1;
	}
	cout<<lo<<endl;
	// your code goes here
	return 0;
}
### Prompt

Create a solution in CPP for 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;

long long n, k;
long long a[200004];

long long f(long long mid){
	long long ret = 0;
	for(int i = 0; i < n; i++){
		if(a[i] == 0){
			if(mid >= 0)	ret += n-1;
			continue;
		}
		if(a[i] < 0){
			long long target = mid/a[i];
			if(target*a[i] > mid)	target++;
			int idx = lower_bound(a, a+n, target)-a;
			ret += n - idx;
			if(idx <= i)	ret -= 1;
		}
		else{
			long long target = mid/a[i];
			if(target*a[i] > mid)	target--;
			int idx = upper_bound(a, a+n, target)-a-1;
			if(idx < 0)	continue;
			ret += idx + 1;
			if(idx >= i)	ret -= 1;
		}
	}
	return ret/2;
}

int main() {
	cin>>n>>k;
	for(int i = 0; i < n; i++)	cin>>a[i];
	sort(a, a+n);
	long long lo = -1e18, hi = 1e18;
	while(lo < hi){
		long long mid = lo + (hi - lo)/2;
		if(f(mid) >= k)	hi = mid;
		else	lo = mid + 1;
	}
	cout<<lo<<endl;
	// your code goes here
	return 0;
}
```