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

题意：
询问区间内异或和刚好为k的字段个数。
思路：
莫队+前缀和。
这个前缀和比较套路，用的是前缀异或和。
字段【l,r】的异或和就是pre[r]^pre[l-1],
这种情况下我们在莫队的过程中记录l，r的pre[i]出现的次数，就可以完成更新了。

注意当L<q[i].l时，要先让记录per[L]出现次数的个数减一。

#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#define fuck(x) cout<<#x<<" = "<<x<<endl;
#define debug(a,i) cout<<#a<<"["<<i<<"] = "<<a[i]<<endl;
#define ls (t<<1)
#define rs ((t<<1)+1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 100086;
const int maxm = 1000086;
const int inf = 2.1e9;
const ll Inf = 999999999999999999;
const int mod = 1000000007;
const double eps = 1e-6;
const double pi = acos(-1);
int num[maxm*2],pre[maxn],a[maxn];

struct node{
    int l,r;
    int id;
}q[maxn];
ll ans[maxn];
ll anss;
int block;

bool cmp(node a,node b){
    if(a.l/block!=b.l/block){return a.l<b.l;}
    return a.r<b.r;
}

int main()
{
    int n,m,k;
    scanf("%d%d%d",&n,&m,&k);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        pre[i]=pre[i-1]^a[i];
    }

    block=sqrt(n);

    for(int i=1;i<=m;i++){
        scanf("%d%d",&q[i].l,&q[i].r);
        q[i].id=i;
    }

    sort(q+1,q+1+m,cmp);
    int L=1,R=0;
    anss=0;
    num[0]=1;
    for(int i=1;i<=m;i++){
        while(L<q[i].l){
            int t=k^pre[L-1];
            num[pre[L-1]]--;///注意语句顺序
            anss-=num[t];
            L++;
        }
        while(R>q[i].r){
            int t=k^pre[R];
            num[pre[R]]--;
            anss-=num[t];
            R--;
        }
        while(L>q[i].l){
            L--;
            int t=k^pre[L-1];
            anss+=num[t];
            num[pre[L-1]]++;
        }
        while(R<q[i].r){
            R++;
            int t=k^pre[R];
            anss+=num[t];
            num[pre[R]]++;
        }
        ans[q[i].id]=anss;
    }
    for(int i=1;i<=m;i++){
        printf("%lld\n",ans[i]);
    }
    return 0;
}

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