Educational Codeforces Round 12 E. Beautiful Subarrays trie求两异或值大于等于k对数

E. Beautiful Subarrays

 

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

 

题意：

　　

　　给你n个数的序列，问你多少区间的异或值大于等于k

　

题解：

　　

　　对于每一个前缀我们插入trie中,

　　询问当前前缀与之前前缀的异或值大于等于k就好

 

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e7+6;
struct Tri
{
    int ch[maxn][2];
    int sz[maxn];
    int tot;
    void init()
    {
        memset(ch,0,sizeof(ch));
        memset(sz,0,sizeof(sz));
        tot=2;
    }
    void insert(int x)
    {
        int u=1;sz[u]++;
        for(int i=30;i>=0;i--)
        {
            int p = (x>>i)&1;
            if(!ch[u][p])ch[u][p]=tot++;
            u=ch[u][p]; sz[u]++;
        }
    }
    int get(int x,int y)
    {
        int u=1;
        long long ans = 0;
        for(int i=30;i>=0;i--)
        {
            int p = (x>>i)&1^1;
            int q = (y>>i)&1;
            if(q==0)ans+=sz[ch[u][p]],u=ch[u][p^1];
            else u=ch[u][p];
            if(u==1) return ans;
        }
        return ans+sz[u];
    }
}T;
int main()
{
    T.init();
    int n,k;
    scanf("%d%d",&n,&k);
    int pre = 0;
    long long ans = 0;
    T.insert(0);
    for(int i=1;i<=n;i++)
    {
        int x;scanf("%d",&x);
        pre^=x;
        ans+=T.get(pre,k);
        T.insert(pre);
    }
    cout<<ans<<endl;
    return 0;
}