Codeforces Round #340 (Div. 2) E. XOR and Favorite Number

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 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:有一个长度为n的数组,有m个查询,问每个查询l,r连续异或和为k的有多少种

分析:设a[i]为前面所有数的异或和,那么[l,r]之间的异或和便为a[l-1]^a[r]

如果知道了[l,r]的异或和为k的种数,接下来便可以在O(1)的时间内得到[l,r-1].[l.r+1],[l-1,r],

[l+1,r]的异或种数,利用莫队算法便能轻松解决

#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define inf -0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define PI acos(-1.0)
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define mem(a, b) memset(a, b, sizeof(a))
typedef long long ll;
const int maxn=101000;
const int MAXM=2000100;
struct node{
    int l,r,id;
}line[maxn];
__int64 ans[maxn],num[MAXM];
int a[maxn];
int unit;

bool cmp(node u,node v){
    if(u.l/unit!=v.l/unit)
        return u.l/unit<v.l/unit;
    return u.r<v.r;
}
int n,m,k;

void solve(){
    int L=1,R=0;
    long long temp=0;
    for(int i=1;i<=m;i++){
        while(R<line[i].r){
            R++;
            temp+=num[a[R]^k];
            num[a[R]]++;
        }
        while(R>line[i].r){
            num[a[R]]--;    //num数值的改变要特别注意
            temp-=num[a[R]^k];
            R--;
        }
        while(L>line[i].l){
            L--;
            temp+=num[a[L]^k];
            num[a[L]]++;
        }
        while(L<line[i].l){
            num[a[L]]--;
            temp-=num[a[L]^k];
            L++;
        }
        ans[line[i].id]=temp;
    }
}

int main(){
    scanf("%d%d%d",&n,&m,&k);
    a[0]=0;
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        a[i]=a[i]^a[i-1];
    }
    for(int i=1;i<=m;i++){
        scanf("%d%d",&line[i].l,&line[i].r);
        line[i].l--;
        line[i].id=i;
    }
    unit=(int)sqrt(n);
    sort(line+1,line+m+1,cmp);
    solve();
    for(int i=1;i<=m;i++)
        printf("%I64d\n",ans[i]);
    return 0;
}