CF 671E XOR and Favorite Number （莫队算法）

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 anda_i of lengthn. Now he asks you to answerm 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 numbers a_i, a_i + 1, ..., a_j is equal to k.

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 integersa_i (0 ≤ a_i ≤ 1 000 000) — Bob's array.

Then m lines follow. Thei-th line contains integersl_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.

Examples

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.

题意：给你n个数，然后询问m次，每次询问一个区间[l,r]，问你在这个区间内有多少子区间，使得子区间内所有值的亦或值为k

分析：

莫队算法：离线处理没有修改的区间查询，当你知道[L,R]的答案后，可以在O(1)的时间内知道[L-1,R]，[L+1,R]，[L,R-1]，[L,R+1]的答案时，就可以使用莫队算法。

用sum[]数组记录前缀亦或值，如果一个区间[l,r]的所有值的亦或值为k，那么sum[r]^sum[l-1]==k，也就是sum[r]^k==sum[l-1]

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define block 320#define maxn 100010#define LL __int64int sum[maxn];int n,m,k;struct node{    int l,r,unit,id;    node(){}    node(int l,int r,int unit,int id): l(l),r(r),unit(unit),id(id) {}    inline bool operator < (const node &x) const    {        if(unit==x.unit) return r<x.r;        return unit<x.unit;    }}E[maxn];LL ans[maxn];LL cnt[maxn*20];int l,r;LL res;void add(int x){    res+=cnt[x^k];    cnt[x]++;}void del(int x){    cnt[x]--;    res-=cnt[x^k];}int main(){    while(scanf("%d%d%d",&n,&m,&k)!=EOF)    {        sum[0]=0;        for(int i=1;i<=n;i++)        {            int x;            scanf("%d",&x);            sum[i]=sum[i-1]^x;        }        for(int i=1;i<=m;i++)        {            scanf("%d%d",&l,&r);            l--;            E[i]=node(l,r,l/block,i);        }        sort(E+1,E+m+1);        memset(cnt,0,sizeof(cnt));        cnt[0]=1;        l=0,r=0,res=0;        for(int i=1;i<=m;i++)        {            while(r<E[i].r)            {                r++;                add(sum[r]);            }            while(r>E[i].r)            {                del(sum[r]);                r--;            }            while(l<E[i].l)            {                del(sum[l]);                l++;            }            while(l>E[i].l)            {                l--;                add(sum[l]);            }            ans[E[i].id]=res;        }        for(int i=1;i<=m;i++) printf("%I64d\n",ans[i]);    }    return 0;}