#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i,n) for(int i=0; i<(n); i++)
int n;
ll k, a[2][200005], c[3], b, l = -1e18, r = 1e18;
int main(){
	scanf("%d%lld", &n, &k);
	rep(i,n){
		scanf("%lld", &b);
		if(b>0) a[0][c[0]++] = b;
		else if(b<0) a[1][c[1]++] = -b;
		else c[2]++;
	}
	rep(i,2) sort(a[i],a[i]+c[i]);
	while(l+1 < r){
		ll m = (l + r) / 2, cnt = 0;
		if(m > 0){
			cnt += c[0] * c[1] + c[2] * (c[0]+c[1]) + c[2] * (c[2]-1) / 2;
			rep(o,2) rep(i,c[o]) cnt += upper_bound(a[o]+i+1,a[o]+c[o],(m-1)/a[o][i]) - (a[o]+i+1);
		}else rep(i,c[0]) cnt += a[1]+c[1] - upper_bound(a[1],a[1]+c[1],(-m)/a[0][i]);
		(cnt < k ? l : r) = m;
	}
	printf("%lld\n", l);
}
### Prompt

Your challenge is to write 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;
using ll = long long;
#define rep(i,n) for(int i=0; i<(n); i++)
int n;
ll k, a[2][200005], c[3], b, l = -1e18, r = 1e18;
int main(){
	scanf("%d%lld", &n, &k);
	rep(i,n){
		scanf("%lld", &b);
		if(b>0) a[0][c[0]++] = b;
		else if(b<0) a[1][c[1]++] = -b;
		else c[2]++;
	}
	rep(i,2) sort(a[i],a[i]+c[i]);
	while(l+1 < r){
		ll m = (l + r) / 2, cnt = 0;
		if(m > 0){
			cnt += c[0] * c[1] + c[2] * (c[0]+c[1]) + c[2] * (c[2]-1) / 2;
			rep(o,2) rep(i,c[o]) cnt += upper_bound(a[o]+i+1,a[o]+c[o],(m-1)/a[o][i]) - (a[o]+i+1);
		}else rep(i,c[0]) cnt += a[1]+c[1] - upper_bound(a[1],a[1]+c[1],(-m)/a[0][i]);
		(cnt < k ? l : r) = m;
	}
	printf("%lld\n", l);
}
```