B. Arpa’s obvious problem and Mehrdad’s terrible solution #383 B

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, ..., an (1 ≤ ai ≤ 105) — the elements of the array.

Output

Print a single integer: the answer to the problem.

Examples

input

2 31 2

output

1

input

6 15 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.

思路：就是在异或回去就好了,考虑一下x为0 的情况

book数组开小了wa了一晚上test11,两个数异或会超过10w,所以以后数组还是尽量开大吧；

#include<iostream>#include<cstdio>#include<cstring>#define ll long long#define N 1000000using namespace std;ll n,x;ll a[N],book[N];int main(){    int i;     scanf("%I64d %I64d",&n,&x);     memset(book,0,sizeof(book)); for(i=1;i<=n;i++) {   scanf("%I64d",&a[i]);   book[a[i]]++;     }     ll w;     ll sum=0;     if(x==0)     { for(i=1;i<=n;i++)       sum+=book[a[i]]-1;      }      else      {    for(i=1;i<=n;i++)           {    w=a[i]^x;               if(book[w])               {   sum+=book[w];   }   }  }  printf("%I64d\n",sum/2);    return 0;}