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.

【题解】



给你n个数字;

m个询问li,ri;

要让你在[li,ri]这个区间里面找到下标对i,j;

使得a[i]xor a[i+1] xor a[i+2]..xor a[j] == k;

让你输出在li,ri内这样的i,j对的个数;

n=10W;

m=100W;

每个数字ai最大为100W为非负数；

设sum[i]表示前i个数字的异或值;

离线处理询问；左端升序排;左端相同右端升序排;

然后从第一个询问开始处理;

设区间为L..R;

则i从L->R递增flag[sum[i]]

然后遇到一个sum[i]则递增答案flag[k^sum[i]];

假设k^sum[i] = sum[x] (x小于i）；

则有sum[x]^sum[i] = k;

而sum[i]^sum[x]实际上就是a[x+1]^a[x+2]..^a[i];

所以这个方式是可行的；

这样我们就能找出所有的点对了；

然后因为询问经过排序处理;

所以相邻询问的l和r和我们刚处理过的L,R是很接近的；

如果l

#include <cstdio>

#include <cmath>

#include <set>

#include <map>

#include <iostream>

#include <algorithm>

#include <cstring>

#include <queue>

#include <vector>

#include <stack>

#include <string>

#define lson L,m,rt<<1

#define rson m+1,R,rt<<1|1

#define LL long long

using namespace std;

const int MAXN = 209999;

const int MAX_SIZE = 1009999;

const int dx[5] = {0,1,-1,0,0};

const int dy[5] = {0,0,0,-1,1};

const double pi = acos(-1.0);

struct abc

{

    int l,r,id;

};

int n,m,k,sum[MAXN];

LL ans[MAXN];

int flag[MAX_SIZE*2] = {0};

abc Q[MAXN];

void input_LL(LL &r)

{

    r = 0;

    char t = getchar();

    while (!isdigit(t)) t = getchar();

    LL sign = 1;

    if (t == '-')sign = -1;

    while (!isdigit(t)) t = getchar();

    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();

    r = r*sign;

}

void input_int(int &r)

{

    r = 0;

    char t = getchar();

    while (!isdigit(t)) t = getchar();

    int sign = 1;

    if (t == '-')sign = -1;

    while (!isdigit(t)) t = getchar();

    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();

    r = r*sign;

}

bool cmp(abc a,abc b)

{

    if (a.l/400!=b.l/400)

        return a.l/400<b.l/400;

    else

        return a.r < b.r;

}

int main()

{

    //freopen("F:\\rush.txt", "r", stdin);

    input_int(n);input_int(m);input_int(k);

    for (int i = 1;i <= n;i++)

    {

        int x;

        input_int(x);

        sum[i] = sum[i-1] ^ x;

    }

    for (int i = 1;i <= m;i++)

        input_int(Q[i].l),input_int(Q[i].r),Q[i].l--,Q[i].id = i;

    sort(Q+1,Q+1+m,cmp);

    int L,R;

    LL s = 0;

    L = Q[1].l,R=Q[1].r;

    for (int i = L;i <= R;i++)

    {

        s+= flag[k^sum[i]];

        flag[sum[i]]++;

    }

    ans[Q[1].id] = s;

    for (int i = 2;i <= m;i++)

    {

        int l = Q[i].l,r=Q[i].r;

        while (L>l)

        {

            L--;

            s+=flag[k^sum[L]];

            flag[sum[L]]++;

        }

        while (L<l)

        {

            flag[sum[L]]--;

            s-=flag[k^sum[L]];

            L++;

        }

        while (R>r)

        {

            flag[sum[R]]--;

            s-=flag[k^sum[R]];

            R--;

        }

        while (R<r)

        {

            R++;

            s+=flag[k^sum[R]];

            flag[sum[R]]++;

        }

        ans[Q[i].id] = s;

    }

    for (int i = 1;i <= m;i++)

        printf("%I64d\n",ans[i]);

    return 0;

}