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

#define li long long int
#define rep(i, to) for (li i = 0; i < ((li)(to)); i++)
#define repp(i, start, to) for (li i = (li)(start); i < ((li)(to)); i++)

li a[200100];
li n;
li k;

inline bool check(li x) {
	li cnt = 0;
	rep(i, n) {
		if (a[i] < 0) {
			cnt += a + n - partition_point(a, a + n, [&](li b) {
				return a[i] * b > x;
			});
		} else {
			cnt += partition_point(a, a + n, [&](li b) {
				return a[i] * b <= x;
			}) - a;
		}
		if (a[i]*a[i] <= x) {
			cnt--;
		}
	}
	cnt /= 2;
	return cnt < k;
}

int main() {
	cin >> n >> k;
	rep(i, n) {
		cin >> a[i];
	}
	sort(a, a + n);

	const li INF = 1ll << 61;

	li left = -INF;
	li right = INF + 1;

	li cnt = 0;
	while (left + 1 < right) {
		li med = (left + right) / 2;
		if (check(med)) {
			left = med;
		} else {
			right = med;
		}
	}

	cout << right << endl;


	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;

#define li long long int
#define rep(i, to) for (li i = 0; i < ((li)(to)); i++)
#define repp(i, start, to) for (li i = (li)(start); i < ((li)(to)); i++)

li a[200100];
li n;
li k;

inline bool check(li x) {
	li cnt = 0;
	rep(i, n) {
		if (a[i] < 0) {
			cnt += a + n - partition_point(a, a + n, [&](li b) {
				return a[i] * b > x;
			});
		} else {
			cnt += partition_point(a, a + n, [&](li b) {
				return a[i] * b <= x;
			}) - a;
		}
		if (a[i]*a[i] <= x) {
			cnt--;
		}
	}
	cnt /= 2;
	return cnt < k;
}

int main() {
	cin >> n >> k;
	rep(i, n) {
		cin >> a[i];
	}
	sort(a, a + n);

	const li INF = 1ll << 61;

	li left = -INF;
	li right = INF + 1;

	li cnt = 0;
	while (left + 1 < right) {
		li med = (left + right) / 2;
		if (check(med)) {
			left = med;
		} else {
			right = med;
		}
	}

	cout << right << endl;


	return 0;
}

```