#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
using ll = long long;
int n;
ll k;
ll a[N];
bool pred(ll m) {
	ll tot = 0;
	for(int i = 0; i < n; ++i) {
		if(a[i] < 0) {
			ll l = -1, r = n;
			while(r - l > 1) {
				ll mid = (l + r) / 2;
				if(a[i] * a[mid] <= m)
					r = mid;
				else
					l = mid;
			}
			tot += (n - r);
		} else {
			ll l = -1, r = n;
			while(r - l > 1) {
				ll mid = (l + r) / 2;
				if(a[i] * a[mid] <= m)
					l = mid;
				else
					r = mid;
			}
			tot += r;
		}
		if(a[i] * a[i] <= m) tot--;
	}
	tot /= 2;
	return tot >= k;
}
int main() {
	scanf("%d %lld", &n, &k);
	for(int i = 0; i < n; ++i)
		scanf("%lld", &a[i]);
	sort(a, a + n);
	// pred(-7);
	// return 0;
	ll lo = -2e18, hi = 2e18;
	while(hi - lo > 1) {
		ll mid = (lo + hi) / 2;
		if(pred(mid))
			hi = mid;
		else
			lo = mid;
	}
	ll ans = hi;
	printf("%lld\n", ans);
	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 int N = 2e5 + 10;
using ll = long long;
int n;
ll k;
ll a[N];
bool pred(ll m) {
	ll tot = 0;
	for(int i = 0; i < n; ++i) {
		if(a[i] < 0) {
			ll l = -1, r = n;
			while(r - l > 1) {
				ll mid = (l + r) / 2;
				if(a[i] * a[mid] <= m)
					r = mid;
				else
					l = mid;
			}
			tot += (n - r);
		} else {
			ll l = -1, r = n;
			while(r - l > 1) {
				ll mid = (l + r) / 2;
				if(a[i] * a[mid] <= m)
					l = mid;
				else
					r = mid;
			}
			tot += r;
		}
		if(a[i] * a[i] <= m) tot--;
	}
	tot /= 2;
	return tot >= k;
}
int main() {
	scanf("%d %lld", &n, &k);
	for(int i = 0; i < n; ++i)
		scanf("%lld", &a[i]);
	sort(a, a + n);
	// pred(-7);
	// return 0;
	ll lo = -2e18, hi = 2e18;
	while(hi - lo > 1) {
		ll mid = (lo + hi) / 2;
		if(pred(mid))
			hi = mid;
		else
			lo = mid;
	}
	ll ans = hi;
	printf("%lld\n", ans);
	return 0;
}


```