Description

XOR is a kind of bit operator, we define that as follow: for two binary base number A and B, let C=A XOR B, then for each bit of C, we can get its value by check the digit of corresponding position in A and B. And for each digit, 1 XOR 1 = 0, 1 XOR 0 = 1, 0 XOR 1 = 1, 0 XOR 0 = 0. And we simply write this operator as ^, like 3 ^ 1 = 2, 4 ^ 3 = 7. XOR is an amazing operator and this is a question about XOR. We can choose several numbers and do XOR operatorion to them one by one, then we get another number. For example, if we choose 2, 3 and 4, we can get 2^3^4=5. Now, you are given N numbers, and you can choose some of them(even a single number) to do XOR on them, and you can get many different numbers. Now I want you tell me which number is the K-th smallest number among them.

Input

First line of the input is a single integer T(T<=30), indicates there are T test cases.

For each test case, the first line is an integer N(1<=N<=10000), the number of numbers below. The second line contains N integers (each number is between 1 and 10^18). The third line is a number Q(1<=Q<=10000), the number of queries. The fourth line contains Q numbers(each number is between 1 and 10^18) K1,K2,......KQ.

Output

For each test case,first output Case #C: in a single line,C means the number of the test case which is from 1 to T. Then for each query, you should output a single line contains the Ki-th smallest number in them, if there are less than Ki different numbers, output -1.

Sample Input

2
2
1 2
4
1 2 3 4
3
1 2 3
5
1 2 3 4 5

Sample Output

Case #1:
1
2
3
-1
Case #2:
0
1
2
3
-1

meaning of the title

Given the length of nn array, there are qq queries, each query by the array subset XOR the result of the kk is small.

thinking

Insert all elements of the array into the linear base, and then rebuild it: enumerate from high to low. For each I I, enumerate J < ij < I. if the jj bit of d[i]d[i] is 11, let it XOR once d[j]d[j] (eliminate the current bit).

After the operation, for d[i]d[i], only the bit ii is 11, and the rest bits are 00. We put the non-zero elements into pp in the order of small to large.

When querying the XOR value of kk, we construct it according to the binary of kk. For bit 11, it is the corresponding p[i]p[i] on XOR. It is correct to think about this construction method carefully.

It should be noted that the result of exclusive or by linear basis is non-zero, and the kk minor difference or value of query may be zero, so we need to judge whether these numbers can exclusive or give 00 when constructing linear basis, and then judge the final result of query.

AC code

#include <bits/stdc++.h>
#define IO                       \
    ios::sync_with_stdio(false); \
    cin.tie(0);                  \
    cout.tie(0);
using namespace std;
typedef long long LL;
const int maxn = 1e5 + 10;
int n, q;

class LinearBase {
public:
    const static int maxn = 64 - 1;
    LL d[maxn], p[maxn], tot;
    void clear() {
        tot = 0;
        memset(d, 0, sizeof(d));
        memset(p, 0, sizeof(p));
    }

    bool insert(LL val) {
        for (int i = maxn - 1; i >= 0; i--) {
            if ((val >> i) & 1) {
                if (!d[i]) {
                    d[i] = val;
                    break;
                } else {
                    val ^= d[i];
                }
            }
        }
        return val > 0;
    }

    LL query_max(LL def = 0) {
        LL res = def;
        for (int i = maxn - 1; i >= 0; i--) {
            res = max(res, res ^ d[i]);
        }
        return res;
    }

    LL query_min() {
        for (int i = 0; i < maxn; i++) {
            if (d[i])
                return d[i];
        }
        return 0;
    }

    void rebuild() {
        memset(p, 0, sizeof(p));
        tot = 0;
        for (int i = maxn - 1; i >= 0; i--) {
            for (int j = i - 1; j >= 0; j--) {
                if ((d[i] >> j) & 1) {
                    d[i] ^= d[j];
                }
            }
        }
        for (int i = 0; i < maxn; i++) {
            if (d[i]) {
                p[tot++] = d[i];
            }
        }
    }

    LL query_kth(LL k) {
        if (k >= 1LL << tot) {
            return -1;
        }
        LL res = 0;
        for (int i = 0; i < maxn; i++) {
            if ((k >> i) & 1) {
                res ^= p[i];
            }
        }
        return res;
    }

} lb;

int main() {
#ifdef LOCAL_IM0QIANQIAN
    freopen("test.in", "r", stdin);
    freopen("test.out", "w", stdout);
#else
    IO;
#endif // LOCAL_IM0QIANQIAN

    int T;
    cin >> T;
    for (int ti = 1; ti <= T; ti++) {
        bool flag = true;
        lb.clear();
        cin >> n;
        for (int i = 0; i < n; i++) {
            LL x;
            cin >> x;
            flag &= lb.insert(x);
        }
        cout << "Case #" << ti << ":" << endl;
        cin >> q;
        lb.rebuild();
        for (int i = 0; i < q; i++) {
            LL x;
            cin >> x;
            cout << lb.query_kth(x - 1 + flag) << endl;
        }
    }
    return 0;
}