There are some beautiful girls in Arpa’s land as mentioned before.

Once Arpa came up with an obvious problem:

Given an array and a number x, count the number of pairs of indices i, j (1 ≤ i < j ≤ n) such that <image>, where <image> is bitwise xor operation (see notes for explanation).

<image>

Immediately, Mehrdad discovered a terrible solution that nobody trusted. Now Arpa needs your help to implement the solution to that problem.

Input

First line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the array and the integer x.

Second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105) — the elements of the array.

Output

Print a single integer: the answer to the problem.

Examples

Input

2 3
1 2


Output

1

Input

6 1
5 1 2 3 4 1


Output

2

Note

In the first sample there is only one pair of i = 1 and j = 2. <image> so the answer is 1.

In the second sample the only two pairs are i = 3, j = 4 (since <image>) and i = 1, j = 5 (since <image>).

A bitwise xor takes two bit integers of equal length and performs the logical xor operation on each pair of corresponding bits. The result in each position is 1 if only the first bit is 1 or only the second bit is 1, but will be 0 if both are 0 or both are 1. You can read more about bitwise xor operation here: <https://en.wikipedia.org/wiki/Bitwise_operation#XOR>.


import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;

public class ArpasobviousproblemandMehrdadsterriblesolution_CodeForces {


	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);
		String [] temp = br.readLine().split(" ");
		int n= Integer.parseInt(temp[0]);
		int x= Integer.parseInt(temp[1]);
		int num []= new int [n];
		int freq[]=new int [10000000+1];
		temp = br.readLine().split(" ");
		for(int i=0;i<n;i++){
			num[i]=Integer.parseInt(temp[i]);
			//num[i]=i%2==0?2:1;
			//num[i]=1;
			freq[num[i]]++;
		}
		BigInteger count=new BigInteger(0+"");
		for(int i=0;i<n;i++){
			int cur = num[i];
			int find = cur^x;
			if(freq[cur]!=0&&find==cur){
				BigInteger b = new BigInteger(freq[cur]+"");
				BigInteger c = new BigInteger(freq[cur]-1+"");
				b=b.multiply(c);
				b=b.divide(new BigInteger(2+""));
				count=count.add(b);
				freq[cur]=0;
			}
			else{
				if(freq[find]!=0 && freq[cur]!=0){
					BigInteger b = new BigInteger(freq[find]+"");
					BigInteger c = b.multiply(new BigInteger(freq[cur]+""));
					count=count.add(c);
					freq[cur]=0;
					freq[find]=0;				
				}
			}
		}
		out.println(count);
		out.flush();
		out.close();


	}
	// the ans 2499950000
	//  my ans 2499950000	
}
