Problem:
Score : 800 points

Problem Statement

Given are N pairwise distinct non-negative integers A_1,A_2,\ldots,A_N.
Find the number of ways to choose a set of between 1 and K numbers (inclusive) from the given numbers so that the following two conditions are satisfied:

The bitwise AND of the chosen numbers is S.

The bitwise OR of the chosen numbers is T.

Constraints

1 \leq N \leq 50

1 \leq K \leq N

0 \leq A_i < 2^{18}

0 \leq S < 2^{18}

0 \leq T < 2^{18}

A_i \neq A_j (1 \leq i < j \leq N)

Input

Input is given from Standard Input in the following format:

N K S T
A_1 A_2 ... A_N

Output

Print the answer.

Sample Input 1

3 3 0 3
1 2 3

Sample Output 1

2

The conditions are satisfied when we choose \{1,2\} or \{1,2,3\}.

Sample Input 2

5 3 1 7
3 4 9 1 5

Sample Output 2

2

Sample Input 3

5 4 0 15
3 4 9 1 5

Sample Output 3

3

Solution in Rust (1.42.0) (judge verdict: Accepted)

use proconio::*;

fn run() {
    input! {
        n: usize,
        k: usize,
        s: usize,
        t: usize,
        a: [usize; n],
    }
    let a: Vec<usize> = a.into_iter().filter(|a| a & s == s && a | t == t).collect();
    if a.is_empty() || s & t != s {
        println!("0");
        return;
    }
    let mut comb = vec![vec![0i64; n + 1]; n + 1];
    comb[0][0] = 1;
    for i in 1..=n {
        comb[i][0] = 1;
        for j in 1..=i {
            comb[i][j] = comb[i - 1][j - 1] + comb[i - 1][j];
        }
    }
    let mut bit = vec![];
    for i in 0..18 {
        if (s >> i) & 1 != (t >> i) & 1 {
            bit.push(i);
        }
    }
    let n = std::cmp::min(n, a.len());
    let k = std::cmp::min(k, n);
    let mut cnt = vec![0; 1 << 18];
    let mut ans = 0i64;
    for b in 0..(1usize << bit.len()) {
        let sign = if b.count_ones() % 2 == 0 {1} else {-1};
        let mut mask = 0;
        for (i, &x) in bit.iter().enumerate() {
            if (b >> i) & 1 == 1 {
                mask |= 1 << x;
            }
        }
        for a in a.iter() {
            cnt[*a & mask] += 1;
        }
        for a in a.iter() {
            let x = *a & mask;
            if cnt[x] == 0 {
                continue;
            }
            let v = cnt[x];
            cnt[x] = 0;
            for c in comb[v].iter().skip(1).take(std::cmp::min(v, k)) {
                ans += sign * *c;
            }
        }
    }
    println!("{}", ans);
}

fn main() {
    run();
}