#include"bits/stdc++.h"
using namespace std;
typedef long long ll;
#define rep(i, n) for(int (i)=0; (i)<(int)(n); (i)++)

ll N, K;
ll A[200000];

// ペア (i, i+1), (i, i+2), ..., (i, N-1) の中の key 以下の数の個数
ll binarySearch(int i, ll key){
	if(A[i] < 0){
		int ok = N;
		int ng = i;
		while(ok - ng > 1){
			int m = (ok+ng) / 2;
			if(A[i]*A[m] <= key) ok = m;
			else ng = m;
		}
		return N-ok;
	}
	else{
		int ok = i;
		int ng = N;
		while(ng - ok > 1){
			int m = (ng+ok) / 2;
			if(A[i]*A[m] <= key) ok = m;
			else ng = m;
		}
		return ok-i;
	}
}

signed main(){
	scanf("%lld%lld", &N, &K);
	rep(i, N){
		scanf("%lld", A+i);
	}
	sort(A, A+N);
	
	// その数以下のペアが K 個以上存在するか？
	// 答えは ok
	ll ng = -1000000000LL*1000000000LL-1LL;
	ll ok = 1000000000LL*1000000000LL+1LL;
	while(ok-ng > 1){
		ll m = (ng+ok) / 2;
		ll c = 0;
		rep(i, N) c += binarySearch(i, m);
		if(c >= K) ok = m;
		else ng = m;
	}
	printf("%lld\n", ok);
}

### 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;
typedef long long ll;
#define rep(i, n) for(int (i)=0; (i)<(int)(n); (i)++)

ll N, K;
ll A[200000];

// ペア (i, i+1), (i, i+2), ..., (i, N-1) の中の key 以下の数の個数
ll binarySearch(int i, ll key){
	if(A[i] < 0){
		int ok = N;
		int ng = i;
		while(ok - ng > 1){
			int m = (ok+ng) / 2;
			if(A[i]*A[m] <= key) ok = m;
			else ng = m;
		}
		return N-ok;
	}
	else{
		int ok = i;
		int ng = N;
		while(ng - ok > 1){
			int m = (ng+ok) / 2;
			if(A[i]*A[m] <= key) ok = m;
			else ng = m;
		}
		return ok-i;
	}
}

signed main(){
	scanf("%lld%lld", &N, &K);
	rep(i, N){
		scanf("%lld", A+i);
	}
	sort(A, A+N);
	
	// その数以下のペアが K 個以上存在するか？
	// 答えは ok
	ll ng = -1000000000LL*1000000000LL-1LL;
	ll ok = 1000000000LL*1000000000LL+1LL;
	while(ok-ng > 1){
		ll m = (ng+ok) / 2;
		ll c = 0;
		rep(i, N) c += binarySearch(i, m);
		if(c >= K) ok = m;
		else ng = m;
	}
	printf("%lld\n", ok);
}

```