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

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

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard 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 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.

题意：给出N个数，找到两个数ai,aj使得ai^aj=X,求出这样的(i,j)有多少对。

坑点：X=0，j>i，结果用longlong，注意异或出的值是否合法

公式要知道a^b=X,a^X=b,b^X=a。

代码

#include<math.h>
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
const int maxn=100005;
int num[maxn];
int result[maxn];
int main()
{
memset(result,0,sizeof(result));
int N,X;
scanf("%d%d",&N,&X);
for(int i=0; i<N; i++)
{
scanf("%d",&num[i]);
result[num[i]]++;
}
long long int result_num=0;
for(int i=0; i<N; i++)
{
if((num[i]^X)==num[i])
result_num+=result[num[i]]-1;
else if((num[i]^X)>=1&&(num[i]^X)<=100000&&result[num[i]^X]>0)
result_num+=result[num[i]^X];
result[num[i]]--;
}
printf("%I64d\n",result_num);
return 0;
}