Mishka and Interesting sum(树状数组前缀和)

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 nelements!

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:

  1. Two integers l and r (1 ≤ l ≤ r ≤ n) are specified — bounds of query segment.
  2. Integers, presented in array segment [l,  r] (in sequence of integers al, al + 1, ..., ar) even number of times, are written down.
  3. 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.

Input

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.

Output

Print m non-negative integers — the answers for the queries in the order they appear in the input.

Examples

Input

3
3 7 8
1
1 3

Output

0

Input

7
1 2 1 3 3 2 3
5
4 7
4 5
1 3
1 7
1 5

Output

0
3
1
3
2

首先,区间出现偶数次的数的异或 = 区间出现过的数的异或 ^ 区间出现奇数次的数的异或。

其次,任何一个数对自身进行偶数次异或都等于0,奇数次异或等于这个数自身。

所以区间出现奇数次的数的异或直接打一个前缀。

对于区间出现过的数的异或,假设一开始直接维护一个[1,n]的大区间,那么在具体位置对于每个数是否出现过我们是不得而知的,也就是没有办法实现状态之间的转移,所以我们动态的维护一个(1,x)的区间,对于每个在x位置出现的数,我们要让它

只对x及以后的区间产生作用,即让每一个出现的数尽可能的往后排。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define MAXN 1000004
#define MOD 1000000009
#define INF 0x7ffffff
#define lowbit(x) (x)&(-x)

using namespace std;

int n,m;
int tree[MAXN];
int pre[MAXN];
int a[MAXN];
int ans[MAXN];
map mp;
vector head[MAXN];
vector index[MAXN];

void add(int k,int num){
    while(k <= n){
        tree[k] ^= num;
        k += lowbit(k);
    }
}

int getPre(int k){
    int sum = 0;
    while(k > 0){
        sum ^= tree[k];
        k -= lowbit(k);
    }
    return sum;
}

int main(){
    while(cin >> n){
        memset(tree,0,sizeof(tree));
        memset(pre,0,sizeof(pre));
        mp.clear();
        for(int i=1;i<=n;++i){
            scanf("%d",&a[i]);
            pre[i] = pre[i-1]^a[i];
        }

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

        for(int i=1;i<=n;++i){
            int x = mp[a[i]];
            if(x){
                add(x,a[i]); //对于之前就出现的过数,再add一次相当于偶数次异或,保证a[i]的位置尽可能靠后
            }
            mp[a[i]] = i;
            add(i,a[i]); //保证a[i]只对此处及之后的区间产生影响,保证了a[i]的位置尽可能靠后
            int len = head[i].size();
            for(int j=0;j

 

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