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 C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>

using namespace std;

int main() {
    int n, k, s, t;
    cin >> n >> k >> s >> t;
    vector<int> a(n);
    for (int i = 0; i < n; ++i)
        cin >> a[i];
    function<vector<tuple<int,int,int>>(int, int)> make = [&](int from, int to) -> vector<tuple<int, int, int>> {
        const uint32_t n = to - from;
        vector<tuple<int, int, int>> res;
        for (uint32_t mask = 0; mask < (1u << n); ++mask) {
            int c = 0;
            uint32_t v = 0xffffffff;
            uint32_t u = 0;
            for (uint32_t i = 0; i < n; ++i) {
                if (mask & (1u << i)) {
                    v = v & a[from + i];
                    u = u | a[from + i];
                    ++c;
                }
            }
            res.emplace_back(v, u, c);
        }
        sort(res.begin(), res.end());
        return res;
    };
    auto left = make(0, n / 2);
    auto right = make(n / 2, n);
    long long count = 0;
    for (auto z: left) {
        int v, u, c;
        tie(v, u, c) = z;
        if ((v & s) != s || (v | s) != v)
            continue;
        if ((u & t) != u || (u | t) != t)
            continue;
        for (auto z1: right) {
            int v1, u1, c1;
            tie(v1, u1, c1) = z1;
            if (c + c1 > k)
                continue;
            if ((v & v1) == s && (u | u1) == t)
                ++count;
        }

    }
    cout << count << "\n";
    return 0;
}