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.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.StringTokenizer;
import java.util.TreeSet;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

public class B {

	public static void main(String[] args) {
		InputReader in = new InputReader(System.in);
		PrintWriter out = new PrintWriter(System.out);
		B solver = new B();
		solver.solve(1, in, out);
		out.close();

	}

	public void solve(int testNumber, InputReader in, PrintWriter out) {

		int n = in.nextInt();
		int x = in.nextInt();
		Integer[] arr = new Integer[n];
		HashMap<Integer, ArrayList<Integer>> map1 = new HashMap<Integer, ArrayList<Integer>>();
		HashMap<Integer, TreeSet<Integer>> map2 = new HashMap<Integer, TreeSet<Integer>>();
		
		for(int i = 0; i < n; i++) {
			arr[i] = Integer.valueOf(in.nextInt());
			if(!map1.containsKey(arr[i])) {
				map1.put(arr[i], new ArrayList<Integer>());
			}
			if(!map2.containsKey(arr[i])) {
				map2.put(arr[i], new TreeSet<Integer>());
			}
			map1.get(arr[i]).add(new Integer(i));
			map2.get(arr[i]).add(new Integer(i));
			
		}
		
		// Count
		long count = 0;
		
		for(int i = 0; i < n; i++) {
			
			Integer needed = arr[i] ^ x;
			
			TreeSet<Integer> set = map2.get(needed);
			if(set == null) continue;
		//	System.out.println(set);
			
			Integer indV = set.ceiling(new Integer(i + 1));
			if(indV == null) continue;
		//	System.out.println(indV);
			
			int ind = Collections.<Integer>binarySearch(map1.get(needed), indV); // map1.get(needed).indexOf(indV);
		//	System.out.println(ind);
			
			count += (map1.get(needed).size() - ind);
			
		}
		
		out.println(count);
		
	}

	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream), 32768);
			tokenizer = null;
		}

		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

		public long nextLong() {
			return Long.parseLong(next());
		}

	}
}
