【Codeforces】665E Beautiful Subarrays

E. 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

Understanding

中文题目传送门见E

Solution

前缀和思想，记录前缀异或值s[i]，考虑贪心，一个数异或上自己为0，对于一段区间[l~r]，异或值=s[l-1]^s[r]，证明:s[l-1]^s[r]=s[l-1]^(s[l-1]^s[l~r])=(s[l-1]^s[l-1])^s[l~r]=s[l~r].

一般异或问题用trie树维护,所以，每次吧s[i]丢入trie树，sz记录当前节点的size，统计答案:将k二进制分解，顺着trie走，当前为0，答案+=sz[1](儿子)else 顺着trie树走下去，记得最后统计=的结果。

// <E.cpp> - Sun Oct  9 19:01:04 2016
// This file is made by YJinpeng，created by XuYike's black technology automatically.
// Copyright (C) 2016 ChangJun High School, Inc.
// I don't know what this program is.

#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#define MOD 1000000007
#define INF 1e9
#define IN inline
#define RG register
using namespace std;
typedef long long LL;
typedef long double LB;
const int MAXN=2e+7;
inline int max(int &x,int &y) {return x>y?x:y;}
inline int min(int &x,int &y) {return x<y?x:y;}
inline LL gi() {
	register LL w=0,q=0;register char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')q=1,ch=getchar();
	while(ch>='0'&&ch<='9')w=w*10+ch-'0',ch=getchar();
	return q?-w:w;
}
int c[MAXN][2],s[MAXN],sz[MAXN];LL ans;int k,tot;
void insert(int x){
    int u=1;
    for(int i=30;i>=0;i--){
        int p=(x>>i)&1;
        if(!c[u][p])c[u][p]=++tot;
        sz[u]++;u=c[u][p];
    }sz[u]++;
}
void src(int x){
    int u=1;
    for(int i=30;i>=0;i--){
        int p=((x>>i)&1)^1;
        if(!((k>>i)&1))ans+=sz[c[u][p]],u=c[u][p^1];
        else u=c[u][p];
    }ans+=sz[u];
}
int main()
{
	freopen("E.in","r",stdin);
	freopen("E.out","w",stdout);
	int n=gi();k=gi();int now=0;tot=1;insert(0);
    while(n--){
        int a=gi();now^=a;
        src(now);insert(now);
    }printf("%lld",ans);
	return 0;
}