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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include "bits/stdc++.h"
using namespace std;

/* macro */
#define rep(i,a,b) for(int i=a;i<b;i++)
#define revrep(i,a,b) for(int i = a; i > b; i--)
#define int long long
#define exist(s,e) ((s).find(e)!=(s).end())
#define all(v) (v).begin(), (v).end()
#define each(s,itr) for(auto (itr) = s.begin(); (itr) != s.end(); (itr)++)
#define sum(v) accumulate(all(v), (0LL))
#define isin(a, b, c) (b <= a && a <= c)
#define println cout << "\n";
#define sz(v) (int)v.size()
#define bin(x) static_cast<bitset<16> >(x)



/* alias */
template<class T> using vec = vector<T>;
typedef vector<int> vi;
typedef pair<int, int> pi;
typedef tuple<int, int, int> ti;
typedef map<int, int> mi;
typedef set<int> si;


/* constant */
const int inf = 1LL << 62;
const int mod = 1e9 + 7;
const int dx[8]={1,0,-1,0,-1,1,-1,1};
const int dy[8]={0,1,0,-1,-1,-1,1,1};
const string alphabet = "abcdefghijklmnopqrstuvwxyz";

/* io_method */
int input(){int tmp;cin >> tmp;return tmp;}
string raw_input(){string tmp;cin >> tmp;return tmp;}
template<class T> void printx(T n){cout << n;}
template<class T, class U> void printx(pair<T, U> p){cout << "(" << p.first << "," << p.second << ")";}
template<class T, class U, class V> void printx(tuple<T, U, V> t){cout << "{" << get<0>(t) << "," << get<1>(t) <<"," << get<2>(t) << "}" << endl;}
template<class T> void printx(vector<T> v){cout << "{";rep(i,0,v.size()){printx(v[i]);if(i != v.size()-1)printx(",");}cout << "}";}
template<class T> void print(T n){printx(n);cout << endl;}
template<class T> void print(set<T> s){cout << "{";each(s, e){if(e != s.begin()) printx(",");printx(*e);}cout << "}" << endl;}
template<class T, class U> void print(map<T, U> mp){cout << "{";each(mp, e){cout << "(" << e -> first << "," << e -> second << ")";}cout << "}" << endl;}
template<class T> void printans(vec<T> v){rep(i,0,sz(v)){cout << v[i] << (i == sz(v) - 1 ? "" : " ");}cout << endl;}


/* general_method */
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
template<class T>T cut(T &a, int l, int r){return T(a.begin()+l, a.begin()+r);}

/* main */

int n, k;
vi a;
vi b;
vi bink;
void in(){
    cin >> n >> k;
    rep(i,0,n){
        a.push_back(input());
        b.push_back(input());
    }
    int t = k;
    while(t){
        bink.push_back(t % 2);
        t /= 2;
    }


}

set<string> hoge;

void dfs(int i, string s, bool flag){

    if(i < 0){
        hoge.insert(s);
        return ;
    }

    if(flag){
        dfs(i - 1, s + '1', true);
    }else{
        if(bink[i] == 1){

            dfs(i - 1, s + '0', true);
            dfs(i - 1, s + '1', false);
        }else{  
            dfs(i - 1, s + '0', false);
        }
    }
}


void solve(){
    
    dfs(sz(bink) - 1, "", false);
    int ans = 0;


    each(hoge, itr){
        string bi = *itr;

        reverse(all(bi));

        int tmp = 0;
        int base = 1;
        rep(i,0,50){
            tmp += (bi[i] == '1') * base;
            base *= 2;
        }

        int val = 0;
        rep(i,0,n){
            if((a[i] | tmp) == tmp) val += b[i];
        }
        chmax(ans, val);
    }

    print(ans);

}

signed main(){

    cin.tie(0);
    ios::sync_with_stdio(false);
    in();
    solve();
    
}