Problem:
Score : 500 points

Problem Statement

Seisu-ya, a store specializing in non-negative integers, sells N non-negative integers. The i-th integer is A_i and has a utility of B_i.
There may be multiple equal integers with different utilities.

Takahashi will buy some integers in this store. He can buy a combination of integers whose bitwise OR is less than or equal to K. He wants the sum of utilities of purchased integers to be as large as possible.

Find the maximum possible sum of utilities of purchased integers.

Constraints

1 \leq N \leq 10^5

0 \leq K < 2^{30}

0 \leq A_i < 2^{30}(1\leq i\leq N)

1 \leq B_i \leq 10^9(1\leq i\leq N)

All input values are integers.

Inputs

Input is given from Standard Input in the following format:

N K
A_1 B_1
:
A_N B_N

Outputs

Print the maximum possible sum of utilities of purchased integers.

Sample Input 1

3 5
3 3
4 4
2 5

Sample Output 1

8

Buy 2 and 3 to achieve the maximum possible total utility, 8.

Sample Input 2

3 6
3 3
4 4
2 5

Sample Output 2

9

Buy 2 and 4 to achieve the maximum possible total utility, 9.

Sample Input 3

7 14
10 5
7 4
11 4
9 8
3 6
6 2
8 9

Sample Output 3

32

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

//四則演算 #pragma GCC target("avx")
//並列計算 #pragma GCC optimize("O3")
//条件分岐を減らす #pragma GCC optimize("unroll-loops")
//浮動小数点演算 #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using VI = vector<int>;
using VL = vector<ll>;
using VS = vector<string>;
template<class T> using PQ = priority_queue<T, vector<T>, greater<T>>;
#define FOR(i,a,n) for(int i=(a);i<(n);++i)
#define eFOR(i,a,n) for(int i=(a);i<=(n);++i)
#define rFOR(i,a,n) for(int i=(n)-1;i>=(a);--i)
#define erFOR(i,a,n) for(int i=(n);i>=(a);--i)
#define each(i, a) for(auto &i : a)
#define SORT(a) sort(a.begin(),a.end())
#define rSORT(a) sort(a.rbegin(),a.rend())
#define fSORT(a,f) sort(a.begin(),a.end(),f)
#define all(a) a.begin(),a.end()
#define out(y,x) ((y)<0||h<=(y)||(x)<0||w<=(x))
#define tp(a,i) get<i>(a)
#define line cout << "-----------------------------\n" 
#define ENDL(i,n) ((i) == (n) - 1 ? "\n" : " ")
#define stop system("pause")
constexpr ll INF = 1000000000;
constexpr ll LLINF = 1LL << 60;
constexpr ll mod = 1000000007;
constexpr ll MOD = 998244353;
constexpr ld eps = 1e-10;
constexpr ld pi = 3.1415926535897932;
template<class T>inline bool chmax(T& a, const T& b) { if (a < b) { a = b; return true; }return false; }
template<class T>inline bool chmin(T& a, const T& b) { if (a > b) { a = b; return true; }return false; }
inline void init() { cin.tie(nullptr); cout.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(15); }
template<class T>inline istream& operator>>(istream& is, vector<T>& v) { for (auto& a : v)is >> a; return is; }
template<class T>inline istream& operator>>(istream& is, deque<T>& v) { for (auto& a : v)is >> a; return is; }
template<class T, class U>inline istream& operator>>(istream& is, pair<T, U>& p) { is >> p.first >> p.second; return is; }
template<class T>inline vector<T> vec(size_t a) { return vector<T>(a); }
template<class T>inline vector<T> defvec(T def, size_t a) { return vector<T>(a, def); }
template<class T, class... Ts>inline auto vec(size_t a, Ts... ts) { return vector<decltype(vec<T>(ts...))>(a, vec<T>(ts...)); }
template<class T, class... Ts>inline auto defvec(T def, size_t a, Ts... ts) { return vector<decltype(defvec<T>(def, ts...))>(a, defvec<T>(def, ts...)); }
template<class T>inline void print(const T& a) { cout << a << "\n"; }
template<class T, class... Ts>inline void print(const T& a, const Ts&... ts) { cout << a << " "; print(ts...); }
template<class T>inline void print(const vector<T>& v) { for (int i = 0; i < v.size(); ++i)cout << v[i] << (i == v.size() - 1 ? "\n" : " "); }
template<class T>inline void print(const vector<vector<T>>& v) { for (auto& a : v)print(a); }
inline string reversed(const string& s) { string t = s; reverse(all(t)); return t; }

int main() {
    init();

    int n, k; cin >> n >> k;
    VI a(n);
    VL b(n);
    FOR(i, 0, n)cin >> a[i] >> b[i];
    print("ffff");

    return 0;
}