#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("O3")
#define ll long long
#define ull unsigned long long
#define db(x) cout<<#x"=["<<(x)<<"]"<<endl
#define CL(a,b) memset(a,b,sizeof(a))
#define fast() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define fr0(i,m) for(int i=0;i<m;i++)
#define fr1(i,m) for(int i=1;i<=m;i++)
//author:fridayfang
//date:19 7月 07
const double esp=1e-8;
const int mod=1e9+7;
const double pi=acos(-1);
const int inf=0x3f3f3f3f;
const int maxn = 1e4 + 5;
const int maxm = 1e6+5;
ll a[maxn];
ll p[64];
int to[100];//
ll t,n,q;
int getBase(){//处理处理出线性基
    CL(p,0);//先要清0
    for(ll i=1;i<=n;i++){
        for(int j=63;j>=0;j--){
            if((a[i]>>j)&1){
                if(!p[j]){p[j] = a[i]; break;}
                a[i]^=p[j];
            }
        }
    }
    // 更小的线性基
    int cnt = 0;
    for(int j=0;j<=63;j++){
        if(!p[j]) continue;
        for(int i=j+1;i<=63;i++){
            if((p[i]>>j)&1) p[i] = p[i]^p[j];
        }
        to[cnt++]=j;
    }
    return cnt;//cnt是线性基的个数 to[0]...to[cnt-1]有正确的映射
}
ll getK(ll k,int bound){//除0外的最k小
    int t = 0;
    ll tmp = k;
    ll ans = 0;
    while(tmp){
        if(t>=bound) return -1;
        if(tmp&1) ans^=p[to[t]];
        tmp = tmp>>1;
        t++;
    }
    return ans;
}



int main(){
    fast();cin>>t;
    for(ll i=1;i<=t;i++){
        cout<<"Case #"<<i<<":\n";
        cin>>n;
        for(ll k=1;k<=n;k++) cin>>a[k];
        int cnt = getBase();
        int t = (cnt==(int)n)?0:1;//不等表示有冗余，则可表示0；此时第k小时第k-1小
        cin>>q;
        ll k;
        for(ll s=1;s<=q;s++){
            cin>>k; k-=t;
            cout<<getK(k,cnt)<<endl;
        }
    }

    
    return 0;
}