codeforces 617E (莫队算法)

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 31 2 1 1 0 31 63 5

output

70

input

5 3 11 1 1 1 11 52 41 3

output

944

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:求询问区间内有多少i,j满足a[i]^a[i+1]^...^a[j]=k.

本题可以再O(1)时间内由[l,r]的结果得到[l+1,r],[l-1,r],[l,r-1]以及[l,r+1]的结果,

所以可以用莫队算法.

主要就是add函数和delete函数,数字不是很大直接开一个数组存下来就好了.

#include <bits/stdc++.h>using namespace std;#define maxn 2111111int pos[111111];int n, m, k;long long cnt[maxn], ans[111111];struct node {    int l, r, id;    bool operator < (const node &a) const {        return pos[l] < pos[a.l] || (pos[l] == pos[a.l] && r < a.r);    }}p[111111];long long a[111111], cur;void add (int pos) {    cur += cnt[a[pos]^k];    cnt[a[pos]]++;}void del (int pos) {    cnt[a[pos]]--;    cur -= cnt[a[pos]^k];}int main () {    //freopen ("in.txt", "r", stdin);    scanf ("%d%d%d", &n, &m, &k);    for (int i = 1; i <= n; i++) {        scanf ("%lld", &a[i]);        if (i > 1) a[i] ^= a[i-1];    }    int block = ceil (sqrt (n*1.0));    for (int i = 1; i <= n; i++) pos[i] = (i-1)/block;    for (int i = 0; i < m; i++) {        scanf ("%d%d", &p[i].l, &p[i].r);        p[i].id = i;    }    sort (p, p+m);    int l = 1, r = 0;    cur = 0;    memset (cnt, 0, sizeof cnt);    cnt[0] = 1;    for (int i = 0; i < m; i++) {        while (r > p[i].r) {            del (r);            r--;        }        while (r < p[i].r) {            r++;            add (r);        }        while (l > p[i].l) {            l--;            add (l-1);        }        while (l < p[i].l) {            del (l-1);            l++;        }        ans[p[i].id] = cur;    }    for (int i = 0; i < m; i++) {        printf ("%lld\n", ans[i]);    }    return 0;}