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: Runtime Error)

#define DEBUG
#define prob CandyAddict
#ifdef  DEBUG
#define MAXN 100300
#else
#define MAXN 100300
#endif

#include <bits/stdc++.h>

using namespace std;
#define rep(i, l, r) for (int i = l; i<r;i++)
#define vi vector<int>
#define all(v) v.begin(),v.end()
#define pb(c) push_back(c)
#define vii vector<pair<int,int>>
#define vvi vector<vector<int>>
#define ii pair<int,int>
#define mp(i, j) make_pair(i,j)
#define ull unsigned long long int
#define ll  long long int
#define endl '\n'
#define PI  3.141592653589793238
#define MOD (1000000007)
#define INF (1100000000)
#define EPS 0.00000001

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
#ifdef DEBUG
    freopen("input", "r", stdin);//use flush(cout)
#endif
    ll n, k;
    cin >> n >> k;
    ll ans = 0;
    ll a[n], b[n];
    rep (i, 0, n) {
        cin >> a[i] >> b[i];
    }
    int lastBit = -1;
    rep (i,0,32)
        if (k&(1<<i)) lastBit = i;

    int dp[n] = {0};
    for (int hole = -1; hole <= lastBit; hole++) {
        ll res = 0;
        int orr =0;
        rep (i, 0, n) {

            bitset<32> mask;
            mask.set();
            mask = mask << (hole + 1);
            ull at = a[i] & mask.to_ulong();
            ull kt = k & mask.to_ulong();

            if (((a[i] & (1 << hole) )== 0) && (at | kt) == kt) {
                res += b[i];
                orr |= a[i];
            }


        }
        cout<<orr<<" "<<k<<endl;
        ans = max(ans, res);



    }

    cout<<ans;
    return 0;
}