Little Mishka enjoys programming. Since her birthday has just passed, her friends decided to present her with array of non-negative integers a1, a2, ..., an of n elements!
Mishka loved the array and she instantly decided to determine its beauty value, but she is too little and can't process large arrays. Right because of that she invited you to visit her and asked you to process m queries.
Each query is processed in the following way:
- Two integers l and r (1 ≤ l ≤ r ≤ n) are specified — bounds of query segment.
- Integers, presented in array segment [l, r] (in sequence of integers al, al + 1, ..., ar) even number of times, are written down.
- XOR-sum of written down integers is calculated, and this value is the answer for a query. Formally, if integers written down in point 2 are x1, x2, ..., xk, then Mishka wants to know the value
, where 
— operator of exclusive bitwise OR.
Since only the little bears know the definition of array beauty, all you are to do is to answer each of queries presented.
The first line of the input contains single integer n (1 ≤ n ≤ 1 000 000) — the number of elements in the array.
The second line of the input contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — array elements.
The third line of the input contains single integer m (1 ≤ m ≤ 1 000 000) — the number of queries.
Each of the next m lines describes corresponding query by a pair of integers l and r (1 ≤ l ≤ r ≤ n) — the bounds of query segment.
Print m non-negative integers — the answers for the queries in the order they appear in the input.
3
3 7 8
1
1 3
0
7
1 2 1 3 3 2 3
5
4 7
4 5
1 3
1 7
1 5
0
3
1
3
2
In the second sample:
There is no integers in the segment of the first query, presented even number of times in the segment — the answer is 0.
In the second query there is only integer 3 is presented even number of times — the answer is 3.
In the third query only integer 1 is written down — the answer is 1.
In the fourth query all array elements are considered. Only 1 and 2 are presented there even number of times. The answer is 
.
In the fifth query 1 and 3 are written down. The answer is 
.
题解：
考虑异或的性质，那么，对于一个区间出现偶数次数的异或和，等价于于求该区间的异或和ˆ该区间所有不同数字的异或和（因为一个数异或奇数次等于原数，异或偶数次为0，你把该区间异或和异或上所有不同数字异或和，相当于所有数字出现次数加了1次后再求区间异或和）
而求区间异或和，我们可以像前缀和那样，先预处理一个前缀异或和，对于每个询问，剩下的问题就是如何搞该区间所有不同数字的异或和
在线不好做，所以可以先离线询问，按区间右端点排序
离散化后，考虑使用一个last数组记录每个数字最后的出现位置。开一个树状数组，再从左到右扫一遍，对于一个数a[i]，若last[a[i]]==0，则直接加入，反之，则证明之前a[i]出现过，再异或一次（即取消该last[a[i]]），然后遇到一个询问的右端点就处理该询问的答案
代码如下：
#include<bits/stdc++.h> #define MAXN 1000005 using namespace std; struct qz{ int l,r,ans,num; }q[MAXN]; int n,m,pt=1; int arr[MAXN],sxor[MAXN],node[MAXN],last[MAXN],tp[MAXN]; bool cmp(qz a,qz b) { return a.r<b.r; } bool cmp1(qz a,qz b) { return a.num<b.num; } int lowbit(int a) { return a&(-a); } void add(int pos,int val) { for(;pos<=n;pos+=lowbit(pos)) node[pos]^=val; } int query(int l,int r) { int ret=0; l--; for(;l!=0;l-=lowbit(l)) ret^=node[l]; for(;r!=0;r-=lowbit(r)) ret^=node[r]; return ret; } int main() { memset(last,0,sizeof(last)); scanf("%d",&n); sxor[0]=0; for(int i=1;i<=n;i++) { scanf("%d",&arr[i]); tp[i]=arr[i]; sxor[i]=sxor[i-1]^arr[i]; } sort(tp+1,tp+1+n); scanf("%d",&m); for(int i=1;i<=m;i++) { scanf("%d%d",&q[i].l,&q[i].r); q[i].num=i; } sort(q+1,q+1+m,cmp); for(int i=1;i<=m;i++) { while(pt<=q[i].r) { int r=lower_bound(tp+1,tp+1+n,arr[pt])-tp; if(last[r]!=0) add(last[r],arr[pt]); last[r]=pt; add(pt,arr[pt]); pt++; } q[i].ans=query(q[i].l,q[i].r)^sxor[q[i].l-1]^sxor[q[i].r]; } sort(q+1,q+1+m,cmp1); for(int i=1;i<=m;i++) printf("%d ",q[i].ans); return 0; }