#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
#define ll long long
using namespace std;
const int N=1e5+10,M=2e6+10,inf=1e9+10;
int n,m,q,a[N],b[N],p[N][22];
int bb[M],nn;
ll mi[23],sa[N];
char ans[M];
vector<pair<int,int> > qu[N];
int tot,rt[N],t[M<<5],L[M<<5],R[M<<5];
int update(int pre,int l,int r,int p){
    int x=++tot;t[x]=t[pre]+1;L[x]=L[pre];R[x]=R[pre];
    if(l==r) return x;
    int mid=(l+r)>>1;
    if(p<=mid) L[x]=update(L[pre],l,mid,p);
    else R[x]=update(R[pre],mid+1,r,p);
    return x;
}
int query(int x,int l,int r,int R){
    if(!x) return 0;
    if(r<=R) return t[x];
    int mid=(l+r)>>1,res=query(::L[x],l,mid,R);
    if(mid<R) res+=query(::R[x],mid+1,r,R);
    return res;
}
int main(){
    // freopen("grid.in","r",stdin);
    // freopen("grid.out","w",stdout);
    scanf("%d%d",&n,&m);
    mi[0]=1;rep(i,1,18) mi[i]=mi[i-1]*10;
    rep(i,1,n) scanf("%d",a+i),bb[++nn]=a[i];
    rep(i,1,m){
        scanf("%d",b+i);
        rep(j,0,18)
            bb[++nn]=p[i][j]=min((ll)inf,(mi[j]/b[i])+((mi[j]%b[i])?1:0));
    }
	sort(bb+1,bb+nn+1);
	nn=unique(bb+1,bb+nn+1)-bb-1;
	rep(i,1,n) a[i]=lower_bound(bb+1,bb+nn+1,a[i])-bb;
	rep(i,1,m){
		rt[i]=rt[i-1];
		rep(j,0,18) p[i][j]=lower_bound(bb+1,bb+nn+1,p[i][j])-bb,rt[i]=update(rt[i],1,nn,p[i][j]);
	}
    rep(i,1,n) sa[i]=sa[i-1]+query(rt[m],1,nn,a[i]);
    scanf("%d",&q);
    rep(i,1,q){
        ll x;scanf("%lld",&x);
        int p=lower_bound(sa+1,sa+n+1,x)-sa;
        if(p>n) {ans[i]='x';continue;}
        qu[p].push_back({x-sa[p-1],i});
    }
    tot=0;
    rep(i,1,n){
        for(auto I:qu[i]){
            int l=1,r=m,j=1,k=I.first,id=I.second;
            while(l<=r){
                int mid=(l+r)>>1;
                if(query(rt[mid],1,nn,a[i])>=k) j=mid,r=mid-1;
                else l=mid+1;
            }
            ll val=1ll*bb[a[i]]*b[j];k-=query(rt[j-1],1,nn,a[i]);
            per(j,18,0) if(val>=mi[j]) {ans[id]=(val/mi[j-k+1])%10+'0';break;}
        }
    }
    rep(i,1,q) putchar(ans[i]);
    return 0;
}