codeforces617e(莫队算法)

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本文介绍了一种针对大量XOR查询的优化算法实现，通过预处理数组并利用分块技术来减少每次查询的时间复杂度。文章详细解释了算法流程，并通过实例展示了其应用效果。

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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.

#include <iostream> 
#include <cmath>  
#include <queue>  
#include <cstdio>  
#include <cstring>  
#include <cstdlib>   
#include <algorithm>  
using namespace std; 
typedef long long ll;
#define maxn 100005
int res[maxn];   //异或操作的前缀和 
int pos[maxn];   //记录i在哪个分块中，用于减小排序的复杂度
ll cnt[1<<20];   //记录某个异或结果有多少个 
ll output_ans[maxn];   //输出结果储存 
int n,m,k;
ll ans=0;
struct Node
{
	int l,r,i;
}q[maxn];
bool cmp(const Node a,const Node b)
{
	return pos[a.l]==pos[b.l]?(a.r<b.r):(pos[a.l]<pos[b.l]);
}
void add(int x)
{
	ans+=cnt[res[x]^k];
	cnt[res[x]]++;
}
void del(int x)
{
	cnt[res[x]]--;
	ans-=cnt[res[x]^k];
}
int main()
{
	while(~scanf("%d%d%d",&n,&m,&k))
	{
		memset(cnt,0,sizeof(cnt));
		memset(res,0,sizeof(ans));
		int a;
		int ss=sqrt(n);
		cnt[0]=1;
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&a);
			res[i]=a^res[i-1];
			pos[i]=i/ss;
		}
		for(int i=1;i<=m;i++)
		{
			scanf("%d%d",&q[i].l,&q[i].r);
			q[i].i=i;
			q[i].l--;
		}
		sort(q+1,q+1+m,cmp);
		int l=0,r=0;
		ans=0;
		for(int i=1;i<=m;i++)
		{
			while(q[i].l<l)
			{
				l--;
				add(l);
			}
			while(q[i].l>l)
			{
				del(l);
				l++;
			}
			while(q[i].r<r)
			{
				del(r);
				r--;
			}
			while(q[i].r>r)
			{
				r++;
				add(r);
			}
			output_ans[q[i].i]=ans;
		}
		for(int i=1;i<=m;i++)printf("%I64d\n",output_ans[i]);
	}
}