Codeforces 340 E. XOR and Favorite Number [莫队算法]

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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

Copy

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.

莫队算法：传说中能够解决一切区间问题的莫队算法。

将区间分块，根据快排序，适合只有询问操作，没有修改操作的题目，可以在o（1）复杂度内得到周围的四个点。

add和del函数：做个前缀和，变成询问 l<=i<=j<=r 满足 pre[j] xor pre[i-1] = k 的 (i,j) 对数，qsc说很显然就能推出来啦；

代码（算是模板）：

#include<bits/stdc++.h>
#define ll long long
using namespace std;

const int maxn = 1<<20;

struct node {
    int l, r, id;
} Q[maxn];

ll n, m, k;
int pos[maxn];
ll flag[maxn], a[maxn], ans[maxn], ANS = 0;




bool cmp(node a, node b)
{
    if(pos[a.l] == pos[b.l]) return a.r < b.r;
    else return pos[a.l] < pos[b.l];
}

int L = 1, R = 0;

void adds(int x)
{
    ANS += flag[a[x]^k];
    flag[a[x]]++;
}

void del(int x)
{
    flag[a[x]]--;
    ANS -= flag[a[x]^k];
}

int main()
{
    while(~scanf("%lld%lld%lld", &n, &m, &k))
    {
        memset(flag, 0, sizeof(flag));
        int sz = sqrt(n);
        for(int i = 1; i <= n; i++)
        {
            scanf("%lld", &a[i]);
            a[i] ^= a[i-1];
            pos[i] = i/sz;
        }
        for(int i = 1; i <= m; i++)
        {
            scanf("%d%d", &Q[i].l, &Q[i].r);
            Q[i].id = i;
        }
        L = 1, R = 0, ANS = 0;
        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--;
                adds(L-1);
            }
            while(R < Q[i].r)
            {
                R++;
                adds(R);
            }
            while(R > Q[i].r)
            {
                del(R);
                R--;
            }
            ans[Q[i].id] = ANS;
        }
        for(int i = 1; i <= m; i++)
            cout << ans[i] << endl;
    }
    return 0;
}