XOR and favorite number (Mo pair algorithm)

E. XOR and favorite numbertime limit per test4 secondsmemory limit per test256 megabytesinputstandard Inputoutputstandard Output

Bob has a favorite numberkanda i of length n. Now he asks your to answer m queries. Each query was given by a pair li and Ri and asks you to count th E number of pairs of integers i and J, such thatl? ≤? I? ≤? j. ≤? r and the xor of the numbers ai,? Ai? +?1,?...,? AJ 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 ai (0?≤? Ai? ≤?1?000?000)-bob ' s array.

Then m lines follow. The i-th line contains integers li and RI (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.

Examplesinputcopy

6 2 3
1 2 1 1 0 3
1 6
3 5

Outputcopy

7
0

Inputcopy

5 3 1
1 1 1) 1 1
1 5
2 4
1 3

Outputcopy

9
4
4

Note

In the first sample the suitable pairs of i and J for the first query is: (1, 2), ( 1, 4), (1, 5), ( 2,3 ), (3, 6), (5, 6), (6, 6). Not a single of these pairs are suitable for the second query.

In the second, the sample XOR equals 1 for any subarrays of an odd length.

Test instructions: There are n number and M inquiry, each inquiry will have a L and R, indicating the interval of inquiry,

Ask how many consecutive sub-ranges are in this interval or K

Suppose we now have a prefix XOR or array sum[], now we ask for the XOR value of the interval [l,r],

Using the sum array is sum[l-1]^sum[r]==k, or k^sum[r]==sum[l-1.

1#include <bits/stdc++.h>2 using namespacestd;3typedefLong LongLL;4 Const intMAXN = 2e6 +Ten;5 intN, M, K, L, R, SZ, A[MAXN];6 LL Sum[maxn], ans, ans[maxn];7 structNode {8     intL, R, id;9 node () {}TenNodeintLintRintID): L (L), R (r), ID (ID) {} One     BOOL operator< (ConstNode & A)Const { A         if(L/sz = = A.l/sz)returnR <A.R; -         returnL <A.L; -     } the } QU[MAXN]; - voidAddintx) { -Ans + = sum[a[x] ^K]; -sum[a[x]]++; + } - voidDelintx) { +sum[a[x]]--; AAns-= sum[a[x] ^K]; at } - intMain () { -scanf"%d%d%d", &n, &m, &k); -      for(inti =1; I <= N; i++) { -scanf"%d", &a[i]); -A[i] ^= a[i-1]; in     } -      for(inti =1; I <= m; i++) { toscanf"%d%d", &AMP;QU[I].L, &QU[I].R); +qu[i].l--; -Qu[i].id =i; the     } *SZ = (int) sqrt (n); $Sort (qu +1, Qu + M +1);Panax NotoginsengL =1, R =0; -      for(inti =1; I <= m; i++) { the          while(L > Qu[i].l) Add (--L); +          while(R < QU[I].R) Add (+ +R); A          while(L < QU[I].L) del (l++); the          while(R > Qu[i].r) del (r--); +Ans[qu[i].id] =ans; -     } $      for(inti =1; I <= m; i++) $printf"%lld\n", Ans[i]); -     return 0; -}

XOR and favorite number (Mo pair algorithm)