【CF 617E】 XOR and Favorite Number （Mo's algorithm）

【CF 617E】  XOR and Favorite Number （Mo's algorithm）

E. XOR and Favorite Number

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Bob has a favorite number k and a_i of length n. Now he asks you to answer m queries. Each query is given by a pair l_i and r_i 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 a_i, a_i + 1, ..., a_j 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 a_i (0 ≤ a_i ≤ 1 000 000) — Bob's array.

Then m lines follow. The i-th line contains integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the parameters of the i-th query.

Output

Print m lines, answer the queries in the order they appear in the input.

Sample test(s)

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.

题目大意是找某个区间内连续异或能得到K的区间数

现在先考虑单组区间查询。对于区间[L,R] 可以求个前缀和pre[i] 表示第1～i的数的异或和。这样[L,R]的异或和即为pre[L-1]^pre[R] 换种思路 预存一下pre[L-1]^k 就可以延长它的生命周期。

这样求[L,R]区间内能异或出k的区间数 从L遍历到R 遍历过程中统计遍历过的前缀异或和的个数 当遍历到i时 pre[i]^k出现的次数即为[L,i]中到i的连续的能异或得到k的区间数

这样如果区间向左或向右就可以O(1)的修改 对于多组查询 就可以上莫队算法了

代码如下：

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