Codeforces617E(莫队)

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 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：询问区间[l，r]内有多少个子区间，其亦或和等于k。

思路：莫队，对于区间[a，b]，区间[a，b+1]的ans等于[a，b]的ans加上区间[a，b]内OXR[b+1]^k的个数

　　   对于[a,b]的亦或和，即为XOR[b]^XOR[a-1]

　　　XOR[b]^XOR[a-1] == k   <==>   XOR[b]^k == XOR[a-1]

　　　因此寻找有多少个XOR[a-1]满足XOR[b]^XOR[a-1] == k ，即寻找有多少个XOR[b]^k

　　　使用一个cnt数组记录当前状态下不同区间亦或和的值出现的次数。

//2017-11-14
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 1100000;
const int LEN = 1000;

int n, m, k, L, R, a[N], XOR[N], block[N];
long long ans, ANS[N], cnt[N];
struct Node{
    int l, r, id;
    bool operator<(const Node x) const {
        if(block[l] == block[x.l])
          return r < x.r;
        return block[l] < block[x.l];
    }
}q[N];

void add(int x){
    ans += cnt[XOR[x]^k];
    cnt[XOR[x]]++;
}

void del(int x){
    cnt[XOR[x]]--;
    ans -= cnt[XOR[x]^k];
}

int main()
{
    //freopen("input.txt", "r", stdin);
    while(~scanf("%d%d%d", &n, &m, &k)){
        XOR[0] = 0;
        for(int i = 1; i <= n; i++){
          scanf("%d", &a[i]);
          XOR[i] = XOR[i-1] ^ a[i];
          block[i] = (i-1)/LEN;
        }
        for(int i = 0; i < m; i++){
            scanf("%d%d", &q[i].l, &q[i].r);
            q[i].id = i;
        }
        sort(q, q+m);
        L = 1, R = 0, ans = 0;
        cnt[0] = 1;
        for(int i = 0; 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 = 0; i < m; i++)
          printf("%lld
", ANS[i]);
    }

    return 0;
}