Problem:
Score: 400 points

Problem Statement

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.

Sample Input 1

4 3
3 3 -4 -2

Sample Output 1

-6

There are six ways to form a pair. The products of those pairs are 9, -12, -6, -12, -6, 8.

Sorting those numbers in ascending order, we have -12, -12, -6, -6, 8, 9. The third number in this list is -6.

Sample Input 2

10 40
5 4 3 2 -1 0 0 0 0 0

Sample Output 2

6

Sample Input 3

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

Sample Output 3

448283280358331064

Solution in Rust (1.15.1) (judge verdict: Accepted)

fn read_stdin() -> String {
    use std::io::Read;
    let mut input = String::new();
    std::io::stdin().read_to_string(&mut input).unwrap();
    input
}

fn scan<'a, I: Iterator<Item = &'a str>, T: std::str::FromStr>(tokens: &mut I) -> T {
    tokens.next().unwrap().parse().ok().unwrap()
}

fn rep<T, F: FnMut() -> T>(n: usize, mut f: F) -> Vec<T> {
    (0..n).map(|_| f()).collect()
}

fn main() {
    let input = read_stdin();
    let tokens = &mut input.split_whitespace();
    let n = scan(tokens);
    let k = scan(tokens);
    let a = rep(n, || scan(tokens));
    println!("{}", solve(n, k, a));
}

fn solve(n: usize, k: usize, a: Vec<i64>) -> i64 {
    let mut a = a;
    a.sort();
    let nx = a.iter().filter(|&&ai| ai > 0).count();
    let ny = a.iter().filter(|&&ai| ai < 0).count();
    let x = &a[a.len() - nx..];
    let y = &a[..ny];
    let n_all_prod = (n * n - n) / 2;
    let n_neg_prod = nx * ny;
    let n_pos_prod = (nx * nx - nx) / 2 + (ny * ny - ny) / 2;
    if k <= n_neg_prod {
        solve_negative(k, x, y)
    } else if k > n_all_prod - n_pos_prod {
        solve_positive(k - (n_all_prod - n_pos_prod), x, y)
    } else {
        0
    }
}

fn solve_negative(k: usize, x: &[i64], y: &[i64]) -> i64 {
    let mut a = -1_000_000_000_000_000_001;
    let mut b = -1;
    while a + 1 < b {
        let c = (a + b) / 2;
        let mut count = 0;
        let mut j = 0;
        for i in 0..x.len() {
            while j < y.len() && x[i] * y[j] <= c {
                j += 1;
            }
            count += j;
        }
        if count < k {
            a = c;
        } else {
            b = c;
        }
    }
    b
}

fn solve_positive(k: usize, x: &[i64], y: &[i64]) -> i64 {
    let mut a = 0;
    let mut b = 1_000_000_000_000_000_000;
    while a + 1 < b {
        let c = (a + b) / 2;
        let mut count = 0;
        let mut j = 0;
        for i in (0..x.len()).rev() {
            while j < i && x[i] * x[j] <= c {
                j += 1;
            }
            count += j;
            if i == j {
                count += (i * i - i) / 2;
                break;
            }
        }
        count += (y.len() * y.len() - y.len()) / 2;
        let mut j = 0;
        for i in (0..y.len()).rev() {
            while j < i && y[i] * y[j] > c {
                j += 1;
            }
            count -= j;
            if i == j {
                count -= (i * i - i) / 2;
                break;
            }
        }
        if count < k {
            a = c;
        } else {
            b = c;
        }
    }
    b
}