Codeforces Round #340 (Div. 2) E. XOR and Favorite Number【莫队算法】

E. XOR and Favorite Number

time limit per test4 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

Bob has a favorite number k and ai of length n. Now he asks you to answer m queries. Each query is given by a pair li and ri 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 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 ≤ li ≤ ri ≤ n) — the parameters of the i-th query.

Output

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

Examples

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.

题意：给你一个序列，K次询问，每次给出一个L，R；询问，从L到R之间有多少个子区间满足异或和为K；

分析：题目给的1e6，但是实际异或操作会超过1e6。区间暴力问题，莫队！

Ans和flag注意使用long long int。

时间复杂度：O(n^1.5)

学习资料参考于：bilibili 莫队算法 卿学姐。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<iomanip>
#include<algorithm>
using namespace std;
#define ll long long int
const int maxn = 1 << 20;
int a[maxn]; int n, m, k;
ll ans[maxn], flag[maxn];
int pos[maxn];
struct node {
int l, r;
int id;
}Q[maxn];
bool cmp(node a, node b) {
if (pos[a.l] == pos[b.l])
return a.r < b.r;
return pos[a.l] < pos[b.l];
}
int L = 1, R = 0;
ll Ans = 0;
void add(int x) {
Ans += flag[a[x] ^ k];
flag[a[x]]++;

}
void del(int x) {
flag[a[x]]--;
Ans -= flag[a[x] ^ k];
}
int main()
{
scanf("%d%d%d", &n, &m, &k);
int sz = sqrt(n);
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
a[i] = a[i] ^ a[i - 1];
pos[i] = i / sz;
}
for (int i = 1; i <= m; i++) { //离线操作
scanf("%d%d", &Q[i].l, &Q[i].r);
Q[i].id = i;
}
sort(Q + 1, Q + 1 + m, cmp);
flag[0] = 1;
for (int i = 1; i <= m; i++) {
while (L < Q[i].l) {
del(L - 1);
L++;
}
while (L > Q[i].l) {
L--;
add(L - 1);
}
while (R < Q[i].r) {
R++;
add(R);
}
while (R > Q[i].r) {
del(R);
R--;

}
ans[Q[i].id] = Ans;
}
for (int i = 1; i <= m; i++) {
printf("%lld\n", ans[i]);
}
return 0;
}