Problem Description

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.

题意：一个长度为 n 的数字序列，m 次询问，以及一个数 k，每次询问给出一个区间 [l,r]，求这个区间内有多少个子区间的所有数异或为 k

思路：

问题实质是求区间 [l,r] 内异或值等于 k 的个数，可转为前缀和来做

在输入过程中令 a[i]^=a[i-1]，那么在区间 [l,r] 内，所有数的异或=k 时，有：a[r]^k=a[l-1]

最后再用莫队处理区间，分块枚举即可

Source Program

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #define PI acos(-1.0) #define E 1e-9 #define INF 0x3f3f3f3f #define LL long long const int MOD=10007; const int N=2000000+5; const int dx[]= {-1,1,0,0}; const int dy[]= {0,0,-1,1}; using namespace std;struct Node{int l,r;//询问的左右端点int id;//询问的编号 }q[N]; int n,m,k,a[N]; int block;//分块 LL ans,cnt[N]; LL res[N];bool cmp(Node a,Node b){//奇偶性排序return (a.l/block)^(b.l/block)?a.l<b.l:(((a.l/block)&1)?a.r<b.r:a.r>b.r); }void add(int x){//统计新的ans+=cnt[a[x]^k];cnt[a[x]]++; } void del(int x){//减去旧的cnt[a[x]]--;ans-=cnt[a[x]^k]; } int main(){while(scanf("%d%d%d",&n,&m,&k)!=EOF){a[0]=0;for(int i=1;i<=n;++i){scanf("%d",&a[i]);a[i]=a[i-1]^a[i];}for(int i=1;i<=m;i++){scanf("%d%d",&q[i].l,&q[i].r);q[i].id=i;}memset(cnt,0,sizeof(cnt));ans=0;cnt[0]=1;block=sqrt(m*2/3*1.0);//分块，卡常数sort(q+1,q+m+1,cmp);//对询问进行排序int l=1,r=0;//左右指针for(int i=1;i<=m;i++){int ql=q[i].l,qr=q[i].r;//询问的左右端点while(l>ql) {l--;add(l-1);}//[l-1,r]while(l<ql) {del(l-1);l++;}//[l+1,r]while(r<qr) add(++r);//[l,r+1]while(r>qr) del(r--);//[l,r-1]res[q[i].id]=ans;//获取答案}for(int i=1;i<=m;i++)printf("%lld\n",res[i]);}return 0; }

 

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