#include <bits/stdc++.h>
using namespace std;
typedef int64_t ll;

const int N=2e5+5;
const ll INF = ll(1e18)+1;
ll A[N];
int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);

    int n;
    ll k;
    cin >> n >> k;
    for(int i=0; i<n; i++) cin >> A[i];
    sort(A, A+n);

	ll lb = -INF, ub = INF;
	while(ub - lb > 1) {
		ll x = (lb + ub)/2;
		bool ok = [&] {
			ll tot = 0;
			for(int i=0; i<n; i++) {
				int l = -1, r = n;
				if(A[i] < 0) {
					// big ---------x-------> small
					//   l ------c----------> r
					while(r - l > 1) {
						int c = ( l+r )/2;
						if( A[c]*A[i]<x ) r = c;
						else l = c;
					}
					tot += n - r;
				} else {
					// small ---------x-------> big
					//     l ------c----------> r
					while( r-l > 1) {
						int c = ( l+r )/2;
						if( A[c]*A[i]<x ) l=c;
						else r=c;
					}
					tot += r;
				}
				if( A[i]*A[i] < x ) tot--;
			}
			tot/=2;
			return tot<k;
		}();
		// value
		if(ok) lb = x;
		else ub = x;
	}

	cout << lb << endl;
	return 0;
}
### 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 int64_t ll;

const int N=2e5+5;
const ll INF = ll(1e18)+1;
ll A[N];
int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);

    int n;
    ll k;
    cin >> n >> k;
    for(int i=0; i<n; i++) cin >> A[i];
    sort(A, A+n);

	ll lb = -INF, ub = INF;
	while(ub - lb > 1) {
		ll x = (lb + ub)/2;
		bool ok = [&] {
			ll tot = 0;
			for(int i=0; i<n; i++) {
				int l = -1, r = n;
				if(A[i] < 0) {
					// big ---------x-------> small
					//   l ------c----------> r
					while(r - l > 1) {
						int c = ( l+r )/2;
						if( A[c]*A[i]<x ) r = c;
						else l = c;
					}
					tot += n - r;
				} else {
					// small ---------x-------> big
					//     l ------c----------> r
					while( r-l > 1) {
						int c = ( l+r )/2;
						if( A[c]*A[i]<x ) l=c;
						else r=c;
					}
					tot += r;
				}
				if( A[i]*A[i] < x ) tot--;
			}
			tot/=2;
			return tot<k;
		}();
		// value
		if(ok) lb = x;
		else ub = x;
	}

	cout << lb << endl;
	return 0;
}
```