#include <bits/stdc++.h>

using namespace std;

const long long M=2e5 + 10;
const long long M2=1e18 ;
long long n, k, l, r, mid, ll, rr, mmid, cnt, i, a[M];

int main(){
	
	cin >> n >> k;
	
	for (i = 0 ; i < n ; i ++ ) cin >> a [ i ] ;
	
	sort ( a , a + n ) ;
	
	l = -M2 ; r = M2 ;
	
	while ( l + 1 < r ){
		mid = l + r >> 1 ;
		cnt = 0;
		for (i = 0 ; i < n ; i ++ ){
			
			if (a [ i ] < 0){
				
				ll = i ; rr = n;
				
				while (ll + 1 < rr){

					mmid = ll + rr >> 1 ;
					if (a[i] * a[mmid] <= mid) rr = mmid;
					else ll = mmid;
					
				}
				
				cnt += n - rr;
			}
			else {
				ll = i ; rr = n;
				
				while (ll + 1 < rr){

					mmid = ll + rr >> 1 ;
					if (a[i] * a[mmid] <= mid) ll = mmid;
					else rr = mmid;
					
				}
				
				cnt += ll - i;
			}
		}
		if ( cnt < k ) l = mid ;
		else r = mid ;
	}
	
	cout << r ;
	
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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;

const long long M=2e5 + 10;
const long long M2=1e18 ;
long long n, k, l, r, mid, ll, rr, mmid, cnt, i, a[M];

int main(){
	
	cin >> n >> k;
	
	for (i = 0 ; i < n ; i ++ ) cin >> a [ i ] ;
	
	sort ( a , a + n ) ;
	
	l = -M2 ; r = M2 ;
	
	while ( l + 1 < r ){
		mid = l + r >> 1 ;
		cnt = 0;
		for (i = 0 ; i < n ; i ++ ){
			
			if (a [ i ] < 0){
				
				ll = i ; rr = n;
				
				while (ll + 1 < rr){

					mmid = ll + rr >> 1 ;
					if (a[i] * a[mmid] <= mid) rr = mmid;
					else ll = mmid;
					
				}
				
				cnt += n - rr;
			}
			else {
				ll = i ; rr = n;
				
				while (ll + 1 < rr){

					mmid = ll + rr >> 1 ;
					if (a[i] * a[mmid] <= mid) ll = mmid;
					else rr = mmid;
					
				}
				
				cnt += ll - i;
			}
		}
		if ( cnt < k ) l = mid ;
		else r = mid ;
	}
	
	cout << r ;
	
	return 0;
}
```