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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

using ull = unsigned long long;

typedef struct tpl{
  ull a;
  ull b;

  bool operator<( const tpl& right ) const {
    return a == right.a ? b < right.b : a < right.a;
  }
};

template <typename T>
inline std::string to_bin_str(T n) {
  std::string str;
  while (n > 0) {
    str.push_back('0' + (n & 1));
    n >>= 1;
  }
  reverse(str.begin(), str.end());
  return str;
}

int main(void) {
  ull n, k;
  cin >> n >> k;
  cin.ignore();
  tpl c[n];
  //ull a[n], b[n];
  int j = 0; // c[i].b <= k
  for (int i = 0; i < n; i++){
    //cin >> a[i] >> b[i];
    cin >> c[i].a >> c[i].b;
    cin.ignore();
  }

  // 1-finder

  string strk = to_bin_str(k);
  unsigned short klen = strk.length(); // kの長さ
  char cstrk[klen];
  strcpy(cstrk, strk.c_str());
  vector<ull> masks;
  ull mask;
  masks.push_back(k);
  unsigned short keta;
  //cout << "klen = " << klen << endl;
  //cout << "cstrk = " << cstrk << endl;
  for (int i = 0; i < klen; i++){
    if(cstrk[i] == '1'){
      keta = klen - i - 1;
      mask = k - (1 << keta);
      mask |= (1 << keta) - 1;
      //cout << "mask = " << mask << endl;
      masks.push_back(mask);
    }
  }

  for (auto itr = masks.begin(); itr != masks.end(); itr++){
    //cout << *itr << endl;
  }

  vector <ull> anss;
  ull ans = 0;
  ull maxans = 0;
  for (auto itr = masks.begin(); itr != masks.end(); itr++){
    ans = 0;
    for (int m = 0; m < n; m++){
      if(*itr == (*itr | c[m].a)){
        ans+=c[m].b;
        //cout << "ans = " << ans << endl;
        //cout << "cmb = " << c[m].b << endl;
      }
    }
    if(maxans < ans) maxans = ans;
    anss.push_back(ans);
  }
  cout << maxans << endl;

  return 0;
}