Codeforces Round #383 (Div. 2) B. Arpa’s obvious problem and Mehrdad’s terrible solution

B. Arpa’s obvious problem and Mehrdad’s terrible solution

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 , where is bitwise xor operation (see notes for explanation).

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, ..., a_n (1 ≤ a_i ≤ 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. so the answer is 1.

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

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.

/*
感觉不会在上分了！呜呜呜呜，B C两个都炸了int
B最多是5e9 int 是2e9
*/
#include<bits/stdc++.h>
using namespace std;
int vis[2621450];//表示记录
int a[2621450];
int main()
{
    //freopen("C:\Users\acer\Desktop\in.txt","r",stdin);
    int n,x;
    scanf("%d%d",&n,&x);
    memset(vis,0,sizeof vis);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        vis[a[i]]++;
    }
    if(n==1)
    {
        cout<<"0"<<endl;
        return 0;
    }
    int cur=0;
    for(int i=1;i<=n;i++)
    {
        if(vis[(x^a[i])]>0)
        {
            if((x^a[i])==a[i])//两个数相等的
                cur+=vis[(x^a[i])]-1;
            else
                cur+=vis[(x^a[i])];
        }
    }
    printf("%d
",cur/2);
    return 0;
}