#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 7;
const ll inf = 1e18;
ll a[maxn];
int main() {
	int n;
	ll k;
	scanf("%d%lld", &n, &k);
	for(int i = 0; i < n; i++) scanf("%lld", &a[i]);
	sort(a, a + n);
	ll L = -inf, R = inf;
	while(L + 1 < R) {
		ll MID = (L + R) / 2;
		ll cnt = 0;
		for(int i = 0; i < n; i++) {
			if(a[i] < 0) {
				int l = i, r = n;
				while(l + 1 < r) {
					int mid = (l + r) / 2;
					if(a[i] * a[mid] <= MID) r = mid;
					else l = mid;
				}
				cnt += n - r;
			}
			else {
				int l = i, r = n;
				while(l + 1 < r) {
					int mid = (l + r) / 2;
					if(a[i] * a[mid] <= MID) l = mid;
					else r = mid;
				}
				cnt += l - i;
			}
		}
		if(cnt < k) L = MID;
		else R = MID;
	}
	printf("%lld\n", R);
	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;
typedef long long ll;
const int maxn = 2e5 + 7;
const ll inf = 1e18;
ll a[maxn];
int main() {
	int n;
	ll k;
	scanf("%d%lld", &n, &k);
	for(int i = 0; i < n; i++) scanf("%lld", &a[i]);
	sort(a, a + n);
	ll L = -inf, R = inf;
	while(L + 1 < R) {
		ll MID = (L + R) / 2;
		ll cnt = 0;
		for(int i = 0; i < n; i++) {
			if(a[i] < 0) {
				int l = i, r = n;
				while(l + 1 < r) {
					int mid = (l + r) / 2;
					if(a[i] * a[mid] <= MID) r = mid;
					else l = mid;
				}
				cnt += n - r;
			}
			else {
				int l = i, r = n;
				while(l + 1 < r) {
					int mid = (l + r) / 2;
					if(a[i] * a[mid] <= MID) l = mid;
					else r = mid;
				}
				cnt += l - i;
			}
		}
		if(cnt < k) L = MID;
		else R = MID;
	}
	printf("%lld\n", R);
	return 0;
}
```