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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using System.Text;

class TEST{
	static void Main(){
		Sol mySol =new Sol();
		mySol.Solve();
	}
}

class Sol{
	public void Solve(){
		
		long tot = 0;
		int v = 0;
		for(int i=0;i<N;i++){
			tot += B[i];
			v |= A[i];
		}
		
		if(v <= K){
			Console.WriteLine(tot);
			return;
		}
		
		long min = long.MaxValue;
		bool[] del = new bool[N];
		for(int j=30;j>=0;j--){
			int kj = (K >> j) & 1;
			int vj = (v >> j) & 1;
			
//Console.WriteLine("j:{0},kj:{1},vj:{2}",j,kj,vj);
			if(kj == 0 && vj == 0) continue;
			if(kj == 1 && vj == 0){
				//ここまでひきわけにできていれば ここで決着がつく．
				long t = 0;
				for(int i=0;i<N;i++) if(del[i]) t += A[i];
				min = Math.Min(min, t);
			}
			if(kj == 1 && vj == 1){
				//ここまで引き分けにしてきた分プラス今回分で決着をつけられる
				long t = 0;
				for(int i=0;i<N;i++) if(del[i] || ((A[i] >> j) & 1) == 1) t += A[i];
				min = Math.Min(min, t);
			}
			if(kj == 0 && vj == 1){
				//ここは引き分けにして下位で決着をつける
				for(int i=0;i<N;i++) if(((A[i] >> j) & 1) == 1) del[i] = true;
			}
		}
		
		// 完全に引き分けにできた場合もOK
		long t0 = 0;
		for(int i=0;i<N;i++) if(del[i]) t0 += A[i];
		min = Math.Min(min, t0);
		
		Console.WriteLine(tot - min);
		
	}
	int N,K;
	int[] A;
	long[] B;
	public Sol(){
		var d = ria();
		N = d[0]; K = d[1];
		A = new int[N];
		B = new long[N];
		for(int i=0;i<N;i++){
			d = ria();
			A[i] = d[0];
			B[i] = d[1];
		}
	}

	static String rs(){return Console.ReadLine();}
	static int ri(){return int.Parse(Console.ReadLine());}
	static long rl(){return long.Parse(Console.ReadLine());}
	static double rd(){return double.Parse(Console.ReadLine());}
	static String[] rsa(char sep=' '){return Console.ReadLine().Split(sep);}
	static int[] ria(char sep=' '){return Array.ConvertAll(Console.ReadLine().Split(sep),e=>int.Parse(e));}
	static long[] rla(char sep=' '){return Array.ConvertAll(Console.ReadLine().Split(sep),e=>long.Parse(e));}
	static double[] rda(char sep=' '){return Array.ConvertAll(Console.ReadLine().Split(sep),e=>double.Parse(e));}
}