Mishka and Interesting sum

Mishka and Interesting sum

time limit per test

3.5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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:

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

Note

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 .

分析：要求区间内出现偶数次数的异或，本质是求区间内不同数的异或；

　　　因为在保存前缀异或后，再异或一次不同数可得到答案；

　　　所以离线树状数组或线段树，维护每个值最后出现的位置；

　　　如果在线，主席树也可，只是爆内存了，遗憾；

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, ls[rt]
#define Rson mid+1, R, rs[rt]
const int maxn=1e6+10;
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
inline ll read()
{
    ll x=0;int f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,k,t,pos[maxn],a[maxn],b[maxn],sum[maxn],ans[maxn],c[maxn];
vector<pii>op[maxn];
void add(int x,int y)
{
    for(int i=x;i<=n;i+=(i&(-i)))
        c[i]^=y;
}
int get(int x)
{
    int ret=0;
    for(int i=x;i;i-=(i&(-i)))
        ret^=c[i];
    return ret;
}
int main()
{
    int i,j;
    scanf("%d",&n);
    rep(i,1,n)scanf("%d",&a[i]),b[i]=a[i],sum[i]=a[i]^sum[i-1];
    sort(b+1,b+n+1);
    int num=unique(b+1,b+n+1)-b-1;
    rep(i,1,n)a[i]=lower_bound(b+1,b+num+1,a[i])-b;
    scanf("%d",&m);
    rep(i,1,m)scanf("%d%d",&j,&k),op[k].pb(mp(j,i));
    rep(i,1,n)
    {
        if(pos[a[i]])add(pos[a[i]],b[a[i]]);
        add(pos[a[i]]=i,b[a[i]]);
        for(pii x:op[i])ans[x.se]=sum[i]^sum[x.fi-1]^get(i)^get(x.fi-1);
    }
    rep(i,1,m)printf("%d
",ans[i]);
    //system("Pause");
    return 0;
}