Beautiful Subarrays

Beautiful Subarrays

time limit per test

3 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

One day, ZS the Coder wrote down an array of integers a with elements a1,  a2,  ...,  an.

A subarray of the array a is a sequence al,  al  +  1,  ...,  ar for some integers (l,  r) such that 1  ≤  l  ≤  r  ≤  n. ZS the Coder thinks that a subarray of a is beautiful if the bitwise xor of all the elements in the subarray is at least k.

Help ZS the Coder find the number of beautiful subarrays of a!

Input

The first line contains two integers n and k (1 ≤ n ≤ 106, 1 ≤ k ≤ 109) — the number of elements in the array a and the value of the parameter k.

The second line contains n integers ai (0 ≤ ai ≤ 109) — the elements of the array a.

Output

Print the only integer c — the number of beautiful subarrays of the array a.

Examples

input

3 1
1 2 3

output

5

input

3 2
1 2 3

output

3

input

3 3
1 2 3

output

2
分析：trie树，保留每个前缀再异或即可；
代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=2e7+10;
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
inline ll read()
{
    ll x=0;int f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,k,t,ch[maxn][2],sz[maxn],tot;
ll ans;
void insert(int p)
{
    int now=0;
    for(int i=30;i>=0;i--)
    {
        int q=(p>>i)&1;
        if(!ch[now][q])ch[now][q]=++tot;
        now=ch[now][q],sz[now]++;
    }
}
ll get(int p)
{
    int now=0;
    ll ans=0;
    for(int i=30;i>=0;i--)
    {
        int q=(p>>i)&1^1,t=(m>>i)&1;
        if(!t)ans+=sz[ch[now][q]],now=ch[now][q^1];
        else now=ch[now][q];
        if(!now)return ans;
    }
    return ans+sz[now];
}
int main()
{
    int i,j;
    insert(0);
    scanf("%d%d",&n,&m);
    while(n--)
    {
        scanf("%d",&j),k^=j;
        ans+=get(k);
        insert(k);
    }
    printf("%lld
",ans);
    //system("Pause");
    return 0;
}