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 = 1100000;
const int MAXM = 1<<22;
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/400==c.L/400? 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=1;i<=n;i++) scanf("%lld",&sum[i]),sum[i]^=sum[i-1];    
    for(int i=1;i<=m;i++) scanf("%d %d",&a[i].L,&a[i].R),a[i].Id = i,a[i].L--;
    sort(a+1,a+m+1,cmp);
    L=0,R=0,cnt[0]=1,ant=0;
    for(int i=1;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=1;i<=m;i++) printf("%lld\n",ans[i]);
    return 0;
}