#include <bits/stdc++.h>
using namespace std;
using ll = long long;

int main(){
	ll N, K;
	cin >> N >> K;
	vector<ll> A(N);
	for(auto&a : A) cin >> a;
	sort(A.begin(), A.end());

	ll ok = 1e18, ng = -1e18;
	while(ok - ng > 1) {
		ll mid = (ok + ng) / 2;
		ll eq_and_less_than_mid_num = 0;
		for(ll i=0;i<N-1;i++) {
			if(A[i] < 0) {
				// j > iについてA[i]との積がmidよりも小さくなるA[j]の数
				ll left = i;
				ll right = N;
				while(right - left > 1) {
					ll upper_mid = (left + right) / 2;
					if(A[i] * A[upper_mid] <= mid) right = upper_mid;
					else left = upper_mid;
				}
				eq_and_less_than_mid_num += (N - right);
			}else {
				ll left = i;
				ll right = N;
				while(right - left > 1) {
					ll upper_mid = (left + right) / 2;
					if(A[i] * A[upper_mid] <= mid) left = upper_mid;
					else right = upper_mid;
				}
				eq_and_less_than_mid_num += (left - i);
			}

		}
		if(eq_and_less_than_mid_num >= K) ok = mid; else ng = mid;
	}
	cout << ok << endl;
}
### Prompt

Your task is to create 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;

int main(){
	ll N, K;
	cin >> N >> K;
	vector<ll> A(N);
	for(auto&a : A) cin >> a;
	sort(A.begin(), A.end());

	ll ok = 1e18, ng = -1e18;
	while(ok - ng > 1) {
		ll mid = (ok + ng) / 2;
		ll eq_and_less_than_mid_num = 0;
		for(ll i=0;i<N-1;i++) {
			if(A[i] < 0) {
				// j > iについてA[i]との積がmidよりも小さくなるA[j]の数
				ll left = i;
				ll right = N;
				while(right - left > 1) {
					ll upper_mid = (left + right) / 2;
					if(A[i] * A[upper_mid] <= mid) right = upper_mid;
					else left = upper_mid;
				}
				eq_and_less_than_mid_num += (N - right);
			}else {
				ll left = i;
				ll right = N;
				while(right - left > 1) {
					ll upper_mid = (left + right) / 2;
					if(A[i] * A[upper_mid] <= mid) left = upper_mid;
					else right = upper_mid;
				}
				eq_and_less_than_mid_num += (left - i);
			}

		}
		if(eq_and_less_than_mid_num >= K) ok = mid; else ng = mid;
	}
	cout << ok << endl;
}
```