CodeForces - 742B Arpa’s obvious problem and Mehrdad’s terrible solution

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 xoroperation (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：找给定序列中有几对数的亦或的结果等于k

思路： 若a^b=c 则有a^c=b且b^c=a

用cnt数组记录下之前出现过的次数

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <stack>
#include <map>
#include <cmath>
#include <vector>
#define max_ 100010
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
int cnt[max_];
int main(int argc, char const *argv[])
{
	int i,n,k;
	scanf("%d%d",&n,&k);
	ll ans=0;
	for(i=1;i<=n;i++)
	{
		int x;
		scanf("%d",&x);
		int t=x^k;
		if(t<=max_)
		{
			ans+=cnt[t];
			cnt[x]++;
		}
	}
	printf("%lld\n",ans);
	return 0;
}