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.util.StringTokenizer;

public class B383 {
	public static void main(String args[]) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		// long n = Long.parseLong(br.readLine());
		StringTokenizer st = new StringTokenizer(br.readLine());
		int n, x;
		n = Integer.parseInt(st.nextToken());
		x = Integer.parseInt(st.nextToken());
		int[] arr=new int[1000000+1];
		long ans=0;
		st = new StringTokenizer(br.readLine());
		for(int i=0;i<n;i++){
			int y=Integer.parseInt(st.nextToken());
			int z=y^x;
			ans+=arr[z];
			arr[y]++;
		}
		System.out.println(ans);

//		ArrayList<Pair> arr[] = new ArrayList[n];
//		st = new StringTokenizer(br.readLine());
//		for (int i = 0; i < n; i++) {
//			arr[i].add(new Pair(Integer.parseInt(st.nextToken()), i));
//		}
//		Collections.sort(arr);
//
//		for (int i = 0; i < n; i++) {
//			System.out.println(arr[i].val);
//		}
//
//		int ceil = (int) Math.pow(2, Math.ceil(Math.log(x) / Math.log(2)));
//		// System.out.println(ceil);
//		int floor = (int) Math.pow(2, Math.floor(Math.log(x) / Math.log(2)));
//
//		int fI = 0, cI = 0;
//		for (int i = 0; i < n; i++) {
//			if (arr[i].get(val) >= floor) {
//				fI = i;
//				break;
//			}
//		}
//		for (int i = 0; i < n; i++) {
//			if (arr[i].val >= ceil) {
//				cI = i;
//				break;
//			}
//		}
//
//		long cnt = 0;
//		for (int i = 0; i < n; i++) {
//			if (arr[i].val < floor) {
//				for (int j = fI; j < n; j++) {
//					if ((arr[i].val ^ arr[j].val) == x) {
//						cnt++;
//					}
//				}
//			} else if (arr[i].val >= ceil) {
//				for (int j = 0; j < cI; j++) {
//					if ((arr[i].val ^ arr[j].val) == x) {
//						cnt++;
//					}
//				}
//			} else {
//				for (int j = 0; j < fI; j++) {
//					if ((arr[i].val ^ arr[j].val) == x) {
//						cnt++;
//					}
//				}
//				for (int j = cI; j < n; j++) {
//					if ((arr[i].val ^ arr[j].val) == x) {
//						cnt++;
//					}
//				}
//			}
//		}
//		System.out.println(floor + " " + fI + " " + cnt);

	}
}

//class Pair implements Comparable {
//	int val;
//	int index;
//
//	public Pair(int val, int index) {
//		super();
//		this.val = val;
//		this.index = index;
//	}

//
//
//	@Override
//	public int compareTo(Pair p) {
//		// TODO Auto-generated method stub
//		if (val > p.val)
//			return 1;
//		else if (val < p.val)
//			return -1;
//		return 0;
//	}
//
//}