Mishka and Interesting sum
Contest: Codeforces Round 365 (Div. 2)
Difficulty rating: 2100
Tags: data structures

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 $$x_{1} \oplus x_{2} \oplus \ldots \oplus x_{k-1}$$, where $$\bigcirc$$ — 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 format:
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 format:
Print m non-negative integers — the answers for the queries in the order they appear in the input.

Constraints:
Time limit: 3.5 s
Memory limit: 256.0 MB

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 $$1 \oplus 2 = 3$$.

In the fifth query 1 and 3 are written down. The answer is $$1 \oplus 3 = 2$$.

Example input:
3
3 7 8
1
1 3

Example output:
0

Example input:
7
1 2 1 3 3 2 3
5
4 7
4 5
1 3
1 7
1 5

Example output:
0
3
1
3
2

Editorial:
Easy to see, that the answer for query is XOR-sum of all elements in the segment xored with XOR-sum of distinct elements in the segment. XOR-sum of all numbers we can find in O(1) using partial sums. As for the XOR-sum of distinct numbers... Let's solve easier problem.
Let the queries be like "find the number of distinct values in a segment". Let's sort all the queries according to their right bounds and iterate through all elements of our array. We also need to make a list last, where lastvalue is the last position of value on the processed prefix of array. Assume we are at position r. Then the answer for the query in the segment [l,  r] (l ≤ r) is $$\sum_{i=1}^{n} cnt_i$$, where cnti = 1 if lastai = i and 0 otherwise. It's easy to store and update such values in cnt. When moving to the next position we have to make the following assignments: cntlastai = 0, cnti = 1, lastai = i. To get described sum in O(log n) we can use segment tree (or Fenwick tree) instead of standard array.
Now let's turn back to our problem. Everything we have to do is to change assignment cnti = 1 to cnti = ai and count XOR-sum instead of sum. Now we can solve this problem in O(n log n).
P.S.: Also there is a solution with O(n sqrt n) complexity (using Mo algo), but we tried to kill it :D