【codeforces 742B】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.

【题目链接】:http://codeforces.com/contest/742/problem/B

【题解】



用个map记录1..i-1里面x xor a[i]的值有多少个；

则那些数字都能和a[i]进行xor运算取得a[i];

因为 a xor b = c

则c xor b = a

且c xor a = b;

O(n)枚举搞一下就好;



【完整代码】

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#include <string>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
void rel(LL &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t) && t!='-') t = getchar();
    LL sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}
void rei(int &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t)&&t!='-') t = getchar();
    int sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}
const int MAXN = 1e5+100;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
int n,x;
int a[MAXN];
map <int,int> dic;
int main()
{
   // freopen("F:\\rush.txt","r",stdin);
    scanf("%d%d",&n,&x);
    rep1(i,1,n)
        scanf("%d",&a[i]);
    LL ans = 0;
    rep1(i,1,n)
    {
        if (dic[a[i]^x])
            ans+=dic[a[i]^x];
        dic[a[i]]++;
    }
    cout << ans << endl;
    return 0;
}