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.

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.

Print a single integer: the answer to the problem.

2 3
1 2

Output

1

Input

6 1
5 1 2 3 4 1

Output

2

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.

#include<stdio.h>
#include<string.h>
//这个题比较看这个运算了，(a^b)==c ---(a^c)==b---(b^c)==a,怎么变都是可以的
//有题目可知10^5*10^5 > 10^9,所以暴力肯定会超时的
//考虑降低复杂度，只需要10^5,时间和空间是互补的
//需要空间记录数据，

//刚才还是比较蒙啊，感觉把二维的降成一维的有点别扭，还是从题意来理解
//我们找的是对数，也就是不同的，不能反转也算一个
//这个题思路就是告诉我们temp（新进来的）,我们看看它应该和谁（已经出现的）是cp（一对），然后我们看看它配偶的个数，有几个就加几
//然后temp又变成任人选的配偶，在他相同类型上+1
//和暴力是不同的思路，之前我们都是直接从一个数开始，从他往后找，现在是新进来一个，往//前找
typedef long long ll;
const ll Max=1e6;
ll a[Max];//装的是i或j的坐标
int main()
{
memset(a,0,sizeof(a));
int n,k,i;
ll ans=0;
scanf("%d %d",&n,&k);
for(i=1;i<=n;i++)
{
int temp;
scanf("%d",&temp);
ans += a[temp^k];
a[temp]++;
}
printf("%I64d\n",ans);
return 0;
}