Arpa’s obvious problem and Mehrdad’s terrible solution

 

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 xoroperation 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: .

 

sol：询问有多少对i，j满足ai^aj = X,这就是Trie树可以轻松支持的，只要注意下X=0时的情况就可以了

Ps：答案会爆int

#include 
      
       using namespace std;typedef long long ll;inline ll read(){    ll s=0;    bool f=0;    char ch=' ';    while(!isdigit(ch))    {        f|=(ch=='-'); ch=getchar();    }    while(isdigit(ch))    {        s=(s<<3)+(s<<1)+(ch^48); ch=getchar();    }    return (f)?(-s):(s);}#define R(x) x=read()inline void write(ll x){    if(x<0)    {        putchar('-'); x=-x;    }    if(x<10)    {        putchar(x+'0');    return;    }    write(x/10);    putchar((x%10)+'0');    return;}#define W(x) write(x),putchar(' ')#define Wl(x) write(x),putchar('\n')const int N=100005;int n,Num,a[N];struct Zidianshu{    int Trie[N*25][25],cnt;    int End[N*25];    inline void Init()    {        cnt=0;    }    inline void Insert(int Num)    {        int i,Root=0;        for(i=0;i<=23;i++)        {            int oo=(Num&(1<
       
        >i;            if(!Trie[Root][oo]) Trie[Root][oo]=++cnt;            Root=Trie[Root][oo];        }        End[Root]++;    }    inline int Query(int Num,int Want)    {        int i,Root=0;        for(i=0;i<=23;i++)        {            int oo=((Want&(1<
        
         >i)^((Num&(1<
         
          >i);            if(!Trie[Root][oo]) return 0;            Root=Trie[Root][oo];        }        return End[Root];    }}Trie;int main(){    Trie.Init();    int i;    long long ans=0;    R(n); R(Num);    for(i=1;i<=n;i++)    {        a[i]=read();        ans+=1ll*Trie.Query(a[i],Num);        Trie.Insert(a[i]);    }    Wl(ans);    return 0;}/*input2 31 2output1input6 15 1 2 3 4 1output2input10 01 2 1 2 1 2 1 2 1 2output20*/