codeforces 617E 莫队算法

又是玄之又玄的bug改到深夜。

先附上代码和题目吧

题目：

Bob has a favorite number k and a_i of length n. Now he asks you to answer m queries. Each query is given by a pairl_i andr_i and asks you to count the number of pairs of integersi andj, such thatl ≤ i ≤ j ≤ r and the xor of the numbersa_i, a_i + 1, ..., a_j is equal tok.

Input

The first line of the input contains integers n,m andk (1 ≤ n, m ≤ 100 000,0 ≤ k ≤ 1 000 000) — the length of the array, the number of queries and Bob's favorite number respectively.

The second line contains n integers a_i (0 ≤ a_i ≤ 1 000 000) — Bob's array.

Then m lines follow. The i-th line contains integers l_i andr_i (1 ≤ l_i ≤ r_i ≤ n) — the parameters of the i-th query.

Output

Print m lines, answer the queries in the order they appear in the input.

Example

Input

6 2 31 2 1 1 0 31 63 5

Output

70

Input

5 3 11 1 1 1 11 52 41 3

Output

944

Note

In the first sample the suitable pairs of i andj for the first query are: (1,2), (1,4), (1, 5), (2,3), (3,6), (5,6), (6,6). Not a single of these pairs is suitable for the second query.

In the second sample xor equals 1 for all subarrays of an odd length

大致题意是：给你一段区间，让你求L，R区间有多少个子序列的异或值为k。莫队算法的思路http://blog.csdn.net/hzj1054689699/article/details/51866615和http://blog.csdn.net/bossup/article/details/39236275已经讲的非常详细了，我肯定没dalao讲的好啊，就不说了，主要提醒的一点是注意排序要分块，我到现在也不知道为什么不分块就会超时，仍然没有参透。哪天参透了再说吧，哈哈哈，分块大法好，关键时刻把命保

代码如下：

#include <bits/stdc++.h>using namespace std;#define LL long longconst int maxn = 1<<20;int n,m,k,nn;struct node{    int l,r,id;};node Q[maxn];LL ans[maxn],ANS;LL flag[maxn];int sum[maxn];int cmp(node a,node b){    if((a.l/nn) == (b.l/nn))return a.r<b.r;    return (a.l/nn)<(b.l/nn);}void del(int rt){    flag[sum[rt]]--;    ANS-=flag[sum[rt]^k];}void add(int rt){    ANS+=flag[sum[rt]^k];    flag[sum[rt]]++;}int main(){    scanf("%d%d%d",&n,&m,&k);    nn=sqrt(n);    memset(flag,0,sizeof(flag));    sum[0]=0;    for(int i=1;i<=n;i++){        scanf("%d",&sum[i]);        sum[i]=sum[i]^sum[i-1];    }//    for(int i=1;i<=n;i++){//        printf("%d ",sum[i]);//    }//    printf("\n");    for(int i=1;i<=m;i++){        scanf("%d%d",&Q[i].l,&Q[i].r);       // Q[i].l--;        Q[i].id=i;    }    sort(Q+1,Q+m+1,cmp);    memset(ans,0,sizeof(ans));    ANS=0;    int L=1,R=0;    flag[0]=1;    for(int i=1;i<=m;i++){        //Q[i].l--;        while(L<Q[i].l){            del(L-1);            L++;        }        while(L>Q[i].l){            L--;            add(L-1);        }        while(R<Q[i].r){            R++;            add(R);        }        while(R>Q[i].r){            del(R);            R--;        }        ans[Q[i].id]=ANS;    }    for(int i=1;i<=m;i++){        printf("%lld\n",ans[i]);    }}