codeforces742B

Arpa’s obvious problem and Mehrdad’s terrible solution

CodeForces - 742B

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: https://en.wikipedia.org/wiki/Bitwise_operation#XOR.

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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
inline ll read()
{
    ll s=;
    bool f=;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<)+(s<<)+(ch^); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<)
    {
        putchar('-'); x=-x;
    }
    if(x<)
    {
        putchar(x+'');    return;
    }
    write(x/);
    putchar((x%)+'');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=;
int n,Num,a[N];
struct Zidianshu
{
    int Trie[N*][],cnt;
    int End[N*];
    inline void Init()
    {
        cnt=;
    }
    inline void Insert(int Num)
    {
        int i,Root=;
        for(i=;i<=;i++)
        {
            int oo=(Num&(<<i))>>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=;
        for(i=;i<=;i++)
        {
            int oo=((Want&(<<i))>>i)^((Num&(<<i))>>i);
            if(!Trie[Root][oo]) return ;
            Root=Trie[Root][oo];
        }
        return End[Root];
    }
}Trie;
int main()
{
    Trie.Init();
    int i;
    long long ans=;
    R(n); R(Num);
    for(i=;i<=n;i++)
    {
        a[i]=read();
        ans+=1ll*Trie.Query(a[i],Num);
        Trie.Insert(a[i]);
    }
    Wl(ans);
    return ;
}
/*
input
2 3
1 2
output
1

input
6 1
5 1 2 3 4 1
output
2

input
10 0
1 2 1 2 1 2 1 2 1 2
output
20
*/