Codeforces617E XOR and Favorite Number（分块 异或）

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.

Example

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：

给定数列a[],和m个询问 Q(L,R),回答每个询问中有多少对(L<=i<=j<=R) ，使得异或为k。

异或转化为前缀和处理。然后就差不多交给分块处理了。

离线处理里比较好理解的一种，高中就会了，注意这里不多BB了。

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#define ll long long
using namespace std;
const int maxn=100010;
int a[maxn],pre[maxn],num[1<<20],n,m,k,sqrtn;
struct Query{
    int id,l,r;  ll ans;
}q[maxn];
bool cmp(const Query a,const Query b)
{
    if(a.l/sqrtn==b.l/sqrtn) return a.r<b.r;
    return a.l<b.l;
}
bool cmp2(const Query a,const Query b)
{
    return a.id<b.id;
}
int main()
{
     scanf("%d%d%d",&n,&m,&k);
     sqrtn=(int)sqrt(n);
     for(int i=1;i<=n;i++) {
          scanf("%d",&a[i]);
          pre[i]=pre[i-1]^a[i];
     }
     for(int i=0;i<m;i++){
         scanf("%d%d",&q[i].l,&q[i].r);
         q[i].id=i;
     }
     sort(q,q+m,cmp);
     int l=1,r=1;
     num[pre[1]]++;num[0]++;
     ll cur=(a[1]==k?1:0);
     for(int i=0;i<m;i++)
     {
        while(r<q[i].r){
            cur+=num[pre[r+1]^k];
            r++;
            num[pre[r]]++;
        }
        while(l<q[i].l){
            num[pre[l-1]]--;
            cur-=num[pre[l-1]^k];
            l++;
        }
        while(l>q[i].l){
            cur+=num[pre[l-2]^k];
            num[pre[l-2]]++;
            l--;
        }
        while(r>q[i].r){
            num[pre[r]]--;
            cur-=num[pre[r]^k];
            r--;
        }
        q[i].ans=cur;
     }
     sort(q,q+m,cmp2);
     for(int i=0;i<m;i++) printf("%lld
",q[i].ans);
     return 0;
}