Codeforces Round #340 (Div. 2) E 莫队+前缀异或和

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 a_i of length n. Now he asks you to answer m queries. Each query is given by a pair l_i and r_i 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 a_i, a_i + 1, ..., a_j 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 a_i (0 ≤ a_i ≤ 1 000 000) — Bob's array.

Then m lines follow. The i-th line contains integers l_i and r_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 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：给你一个长度为n的序列 q个查询[l,r]  问[l,r] 有多少个子区间的异或和为k

题解：莫队+前缀异或和

 #include<iostream>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #include<map>
 #include<queue>
 #include<stack>
 #include<vector>
 #include<set>
 #define ll __int64
 using namespace std;
 int n,m,k;
 struct node
 {
     int l,r,id;
 }N[];
 int p[];
 int block;
 int a[];
 int x[];
 int mp[];
 ll ans=;
 ll re[];
 int cmp(struct node aa,struct node bb)
 {
     if(p[aa.l]==p[bb.l])
         return  aa.r<bb.r;
     else
         return p[aa.l]<p[bb.l];
 }
 void update(int w,int h)
 {
     if(h==){
         ans=ans+mp[x[w]^k];
         mp[x[w]]++;
     }
     else
     {
         mp[x[w]]--;
         ans=ans-mp[x[w]^k];
     }
 }
 int main()
 {
     scanf("%d %d %d",&n,&m,&k);
     for(int i=;i<=n;i++)
         scanf("%d",&a[i]);
     x[]=;
     mp[]=;
     for(int i=;i<=n;i++)
         x[i]=a[i]^x[i-];
     for(int i=;i<=m;i++){
         scanf("%d %d",&N[i].l,&N[i].r);
         N[i].id=i;
     }
     block=(int)sqrt((double)n);
     for(int i=;i<=n;i++)
         p[i]=(i-)/block+;
     sort(N+,N++m,cmp);
     ans=;
     for(int i=,l=,r=;i<=m;i++)
     {
         for(;r<N[i].r;r++) update(r+,);
         for(;l>N[i].l;l--) update(l-,);// 取异或的原因
         for(;r>N[i].r;r--) update(r,-);
         for(;l<N[i].l;l++) update(l-,-);//
         re[N[i].id]=ans;
     }
     for(int i=;i<=m;i++)
     printf("%I64d\n",re[i]);
    return ;
 }
 /*
 10 2 0
 0 0 0 0 0 0 0 0 0 0
 2 8
 2 8
 */