#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;

ll n, k, a[200005];
ll lt=-1e18, rt=1e18, md, ans, cnt;

ll neg(ll p) {
	ll l=p, r=n, m, x;
	while(l+1!=r) {
		m = l + r >> 1;
		if(a[p] * a[m] <= md) r = m;
		else l = m;
	}
	return n - r;
}

ll pos(ll p) {
	ll l=p, r=n, m, x;
	while(l+1!=r) {
		m = l + r >> 1;
		if(a[p] * a[m] <= md) l = m;
		else r = m;
	}
	return l - p;
}

int main() {
	ll i;
	cin >> n >> k;
	for(i=0; i<n; i++) scanf("%lld", &a[i]);
	sort(a, a+n);
	while(lt <= rt) {
		md = lt + rt >> 1;
		cnt = 0;
		for(i=0; i<n; i++) {
			if(a[i] < 0) {
				cnt += neg(i);
			} else {
				cnt += pos(i);
			}
		}
		if(cnt < k) lt = md + 1;
		else rt = md -1, ans = md;
	}
	cout << ans;
	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 <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;

ll n, k, a[200005];
ll lt=-1e18, rt=1e18, md, ans, cnt;

ll neg(ll p) {
	ll l=p, r=n, m, x;
	while(l+1!=r) {
		m = l + r >> 1;
		if(a[p] * a[m] <= md) r = m;
		else l = m;
	}
	return n - r;
}

ll pos(ll p) {
	ll l=p, r=n, m, x;
	while(l+1!=r) {
		m = l + r >> 1;
		if(a[p] * a[m] <= md) l = m;
		else r = m;
	}
	return l - p;
}

int main() {
	ll i;
	cin >> n >> k;
	for(i=0; i<n; i++) scanf("%lld", &a[i]);
	sort(a, a+n);
	while(lt <= rt) {
		md = lt + rt >> 1;
		cnt = 0;
		for(i=0; i<n; i++) {
			if(a[i] < 0) {
				cnt += neg(i);
			} else {
				cnt += pos(i);
			}
		}
		if(cnt < k) lt = md + 1;
		else rt = md -1, ans = md;
	}
	cout << ans;
	return 0;
}
```