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.*;
import java.util.*;

public class R383B {
	public static void main(String[] args) throws Exception {
		BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
		Scanner s = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		
		String[] split = f.readLine().split("\\s+");
		int n = Integer.parseInt(split[0]), x = Integer.parseInt(split[1]);
		Integer[] a=  new Integer[n];
		long[] freq = new long[1_000_000];
		split = f.readLine().split("\\s+");
		TreeSet<Integer> dup = new TreeSet<>();
		for(int i = 0; i < n; i++) {
			a[i] = Integer.parseInt(split[i]);
			dup.add(a[i]);
			freq[a[i]]++;
		}
//		for(int i = 0; i <n; i++) {
//			System.out.println(a[i] + " " + (a[i]^x) + " " + Integer.toBinaryString(a[i]));
//		}
//		System.out.println();
		int[] A = new int[dup.size()];
		if(dup.size() == 1) {
			if(x == 0) {
				long fr = freq[dup.pollFirst()];
//				System.out.println(fr);
				System.out.println((fr*(fr-1)) / 2);
				return;
			} else {
				System.out.println(0);
				return;
			}
		}
		long res = 0;
		if(x == 0) {
			for(int i = 0; i < freq.length; i++) {
				if(freq[i] > 1) {
					res += (freq[i] * (freq[i]-1)) / 2;
				}
			}
			System.out.println(res);
			return;
		}
		int index = 0;
//		System.out.println(dup.toString());
		boolean[] skip = new boolean[1_000_000];
		while(!dup.isEmpty()) A[index++] = dup.pollFirst();
		int[] it = new int[A.length];
		for(int i = 0; i < A.length; i++) it[i] = A[i]^x;
//		System.out.println(Arrays.toString(it));
		for(int i = 0; i < A.length; i++) {
//			System.out.println(A[i]);
//			System.out.println("! " + A[i] + " " + A[i-1]);
//			if((A[i] ^ A[i-1]) == x){
//				System.out.println(A[i] + " " + A[i-1]);
//				long mF = Math.max(freq[A[i]], freq[A[i-1]]);
//				long mF2 = Math.min(freq[A[i]], freq[A[i-1]]);
//				if(mF == 1 && mF2 == mF) {
//					res += 1;
//				} else 
//				res += mF * mF2;
//			}
			
			if(freq[it[i]] != 0 && !skip[A[i]]) {
				skip[A[i]] = true;
				long mF = Math.max(freq[A[i]], 0);
				long mF2 = Math.max(freq[it[i]], 0);
				if(mF == 1 && mF2 == mF) {
					res += 1;
				} else 
				res += mF * mF2;
//				System.out.printf("%d %d\n", mF, mF2);
			}
		}
		System.out.println(res/2);
	}
}