CodeForces-617E XOR And Favorite Numbers(莫队)

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_jis 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.

题解：首先要知道 a^b=c 等价于 a=c^b;   我们用a[i]记录前I个数的异或和，然后离线处理所有区间（对于所有区间我们按L所在块为第一排序，该询问的r为第二排序，对所有询问区间排序）；

对于新增加的一个数我们加上前区间异或等于k^s的数的个数，然后更新一下异或为s的数量，对于一个需要去掉的数，我需要先新更新这个数的数量，然后减去k^s的数量即可；

离线跑一遍，

参考代码为：

 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<iostream>
 #include<algorithm>
 #define LL long long
 using namespace std;
 const int Max = ;
 const int MAXM = <<;
 struct Point
 {
     int L ,R,Id;
 } a[Max];

 LL sum[Max],ans[Max],k;
 int n,m,L,R;
 LL cnt[MAXM],ant;
 bool cmp(Point b,Point c)
 {
     return b.L/==c.L/? b.R<c.R:b.L<c.L;
 }
 void Dec(LL s)
 {
     --cnt[s];
     ant-=cnt[s^k];
 }
 void Inc(LL s)
 {
     ant += cnt[s^k];
     cnt[s]++;
 }
 int main()
 {
     scanf("%d %d %lld",&n,&m,&k);
     LL data;
     for(int i=;i<=n;i++) scanf("%lld",&sum[i]),sum[i]^=sum[i-];
     for(int i=;i<=m;i++) scanf("%d %d",&a[i].L,&a[i].R),a[i].Id = i,a[i].L--;
     sort(a+,a+m+,cmp);
     L=,R=,cnt[]=,ant=;
     for(int i=;i<=m;i++)
     {
         while(R<a[i].R) Inc(sum[++R]);
         while(R>a[i].R) Dec(sum[R--]);
         while(L<a[i].L) Dec(sum[L++]);
         while(L>a[i].L) Inc(sum[--L]);
         ans[a[i].Id]=ant;
     }
     for(int i=;i<=m;i++) printf("%lld\n",ans[i]);
     return ;
 }