#include <cstdio>
#include <cmath>
#include <algorithm>
#define N 200005
 
int n;
long long a[N];
 
inline long long cnt(long long x) {
    long long ans = 0;
    for (int i = 0; i < n; ++i)
        if (!a[i]) ans += x >= 0 ? n - i - 1 : 0;
        else if (a[i] < 0) ans += std::min(n - i - 1, (int)(a + n - std::lower_bound(a, a + n, (long long)std::ceil((long double)x / a[i]))));
        else if (a[i] > 0) ans += std::max(0, (int)(std::upper_bound(a, a + n, (long long)std::floor((long double)x / a[i])) - a - i - 1));
    return ans;
}
 
int main() {
    long long k;
    scanf("%d%lld", &n, &k);
    for (int i = 0; i < n; ++i) scanf("%lld", a + i);
    std::sort(a, a + n);
    long long l = -1e18, r = 1e18;
    while (l < r) {
        long long mid = l + r >> 1;
        if (cnt(mid) < k) l = mid + 1;
        else r = mid;
    }
    printf("%lld\n", l);
    return 0;
}

### Prompt

Create a solution in cpp for 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 <cstdio>
#include <cmath>
#include <algorithm>
#define N 200005
 
int n;
long long a[N];
 
inline long long cnt(long long x) {
    long long ans = 0;
    for (int i = 0; i < n; ++i)
        if (!a[i]) ans += x >= 0 ? n - i - 1 : 0;
        else if (a[i] < 0) ans += std::min(n - i - 1, (int)(a + n - std::lower_bound(a, a + n, (long long)std::ceil((long double)x / a[i]))));
        else if (a[i] > 0) ans += std::max(0, (int)(std::upper_bound(a, a + n, (long long)std::floor((long double)x / a[i])) - a - i - 1));
    return ans;
}
 
int main() {
    long long k;
    scanf("%d%lld", &n, &k);
    for (int i = 0; i < n; ++i) scanf("%lld", a + i);
    std::sort(a, a + n);
    long long l = -1e18, r = 1e18;
    while (l < r) {
        long long mid = l + r >> 1;
        if (cnt(mid) < k) l = mid + 1;
        else r = mid;
    }
    printf("%lld\n", l);
    return 0;
}

```