# include <bits/stdc++.h>
# define LL long long
using namespace std;
const int N = 2e5 + 10, INF = 1e9 + 100, lg = 19;
LL n, k, a[N];
inline LL solve(LL m){
	LL cur_ans = 0;
	for(int i = 1; i <= n; i++){
		if(a[i] < 0){
			int ll = 1, rr = i - 1, ans = i;
			while(ll <= rr){
				int mm = (ll + rr) >> 1;
				if(a[mm] * a[i] < m){
					rr = mm - 1;
					ans = mm;
				}else{
					ll = mm + 1;
				}
			}
			cur_ans += i - ans;
		}
		if(a[i] >= 0){
			int ll = 1, rr = i - 1, ans = 0;
			while(ll <= rr){
				int mm = (ll + rr) >> 1;
				if(a[mm] * a[i] < m){
					ll = mm + 1;
					ans = mm;
				}else{
					rr = mm - 1;
				}
			}
			cur_ans += ans;
		}
	}
	return cur_ans;
}
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	cin >> n >> k;
	for(int i = 1; i <= n; i++){
		cin >> a[i];
	}
	sort(a + 1, a + n + 1);
	LL l = -1e18, r = 1e18, res = 0;
	while(l <= r){
		LL m = (l + r) / 2, cur_ans = solve(m);
		if(cur_ans < k){
			l = m + 1;
			res = m;
		}else{
			r = m - 1;
		}
	}
	cout << res << endl;
}

### Prompt

In cpp, your task is to solve 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>
# define LL long long
using namespace std;
const int N = 2e5 + 10, INF = 1e9 + 100, lg = 19;
LL n, k, a[N];
inline LL solve(LL m){
	LL cur_ans = 0;
	for(int i = 1; i <= n; i++){
		if(a[i] < 0){
			int ll = 1, rr = i - 1, ans = i;
			while(ll <= rr){
				int mm = (ll + rr) >> 1;
				if(a[mm] * a[i] < m){
					rr = mm - 1;
					ans = mm;
				}else{
					ll = mm + 1;
				}
			}
			cur_ans += i - ans;
		}
		if(a[i] >= 0){
			int ll = 1, rr = i - 1, ans = 0;
			while(ll <= rr){
				int mm = (ll + rr) >> 1;
				if(a[mm] * a[i] < m){
					ll = mm + 1;
					ans = mm;
				}else{
					rr = mm - 1;
				}
			}
			cur_ans += ans;
		}
	}
	return cur_ans;
}
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	cin >> n >> k;
	for(int i = 1; i <= n; i++){
		cin >> a[i];
	}
	sort(a + 1, a + n + 1);
	LL l = -1e18, r = 1e18, res = 0;
	while(l <= r){
		LL m = (l + r) / 2, cur_ans = solve(m);
		if(cur_ans < k){
			l = m + 1;
			res = m;
		}else{
			r = m - 1;
		}
	}
	cout << res << endl;
}

```