莫队算法初识~~CodeForces

E. XOR and Favorite Number

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Bob has a favorite number k and ai of length n. Now he asks you to answer m queries. Each query is given by a pair li and ri and asks you to count the number of pairs of integers i and j, such that l ≤ i ≤ j ≤ r and the xor of the numbers ai, ai + 1, ..., aj is equal to k.

Input

The first line of the input contains integers n, m and k (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 ai (0 ≤ ai ≤ 1 000 000) — Bob's array.

Then m lines follow. The i-th line contains integers li and ri (1 ≤ li ≤ ri ≤ n) — the parameters of the i-th query.

Output

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

Examples

input

6 2 3
1 2 1 1 0 3
1 6
3 5

output

7
0

input

5 3 1
1 1 1 1 1
1 5
2 4
1 3

output

9
4
4

Note

In the first sample the suitable pairs of i and j 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.

 题目链接

http://codeforces.com/problemset/problem/617/E

题意

给你一段长度为n的区间

有m次询问

和一个要求的数K

m次查询区间L~R内有多少对 i j 满足 ai^ai+1......^aj =k;

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1<<20;
/*
莫队算法：
只有查询没有修改的操作
O(1)查询
n^1.5

*/
struct node{
    int l,r,id;
    //  左 右 第几个询问
}Q[maxn];

int pos[maxn];
long long ans[maxn];
long long flag[maxn]; //每个前缀出现的次数
int a[maxn];
bool cmp(node a,node b){
    if(pos[a.l]==pos[b.l])
        return a.r<b.r;
    return pos[a.l]<pos[b.l];      //按块排序
}


int n,m,k;
int L=1,R=0;
long long Ans=0;
void add(int x){
    Ans+=flag[a[x]^k];  //前缀和异或
    flag[a[x]]++;
}
void del(int x){
    flag[a[x]]--;
    Ans-=flag[a[x]^k];   //删去多余的前缀和
}
int main()
{
    scanf("%d%d%d",&n,&m,&k);
    int sz=sqrt(n);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        a[i]=a[i]^a[i-1];  //前缀和
        pos[i]=i/sz;   //块
    }
    for(int i=1;i<=m;i++){
        scanf("%d%d",&Q[i].l,&Q[i].r);
        Q[i].id = i;   //查询顺序
    }
    sort(Q+1,Q+1+m,cmp);
    flag[0]=1;  //每个前缀出现的次数
    for(int i=1;i<=m;i++){
        while(L<Q[i].l){
            del(L-1);
            L++;  //从左往右走
        }
        while(L>Q[i].l){
            L--;
            add(L-1);
        }
        while(R<Q[i].r){
            R++;  //往右走
            add(R);
        }
        while(R>Q[i].r){
            del(R);
            R--;
        }
        ans[Q[i].id]=Ans;   //第i次查询的结果
    }
    for(int i=1;i<=m;i++){
       cout<<ans[i]<<endl;
    }



}