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

int main()
{
	int N;
	long long K;
	cin >> N >> K;
	vector<long long> A( N );
	for( int i = 0; i < N; i++ ) {
		cin >> A[i];
	}

	sort( A.begin(), A.end() );
	long long l = -1e9 * 1e9;
	long long r = 1e9 * 1e9;
	while( r - l > 1 ) {
		long long m = (l + r) / 2;
		long long k = 0;
		for( int i = 0; i < N; i++ ) {
			if( A[i] == 0 ) {
				if( m >= 0 ) k += N - (i + 1);
			}
			else if( A[i] < 0 ) {
				int L = i;
				int R = N;
				while( R - L > 1 ) {
					int M = (L + R) / 2;
					if( A[i] * A[M] <= m ) R = M;
					else L = M;
				}
				k += N - R;
			}
			else {
				int L = i;
				int R = N;
				while( R - L > 1 ) {
					int M = (L + R) / 2;
					if( A[i] * A[M] <= m ) L = M;
					else R = M;
				}
				k += L - i;
			}
		}
		if( k >= K ) r = m;
		else l = m;
	}

	long long ans = r;
	cout << ans << endl;
}

### Prompt

Generate 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;

int main()
{
	int N;
	long long K;
	cin >> N >> K;
	vector<long long> A( N );
	for( int i = 0; i < N; i++ ) {
		cin >> A[i];
	}

	sort( A.begin(), A.end() );
	long long l = -1e9 * 1e9;
	long long r = 1e9 * 1e9;
	while( r - l > 1 ) {
		long long m = (l + r) / 2;
		long long k = 0;
		for( int i = 0; i < N; i++ ) {
			if( A[i] == 0 ) {
				if( m >= 0 ) k += N - (i + 1);
			}
			else if( A[i] < 0 ) {
				int L = i;
				int R = N;
				while( R - L > 1 ) {
					int M = (L + R) / 2;
					if( A[i] * A[M] <= m ) R = M;
					else L = M;
				}
				k += N - R;
			}
			else {
				int L = i;
				int R = N;
				while( R - L > 1 ) {
					int M = (L + R) / 2;
					if( A[i] * A[M] <= m ) L = M;
					else R = M;
				}
				k += L - i;
			}
		}
		if( k >= K ) r = m;
		else l = m;
	}

	long long ans = r;
	cout << ans << endl;
}

```