#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long ll;
typedef double db;
inline int read()
{int x=0,f=1; char ch=getchar();while(ch<'0'||ch>'9') { if(ch=='-') f=-1; ch=getchar(); }while(ch>='0'&&ch<='9') { x=(x<<1)+(x<<3)+(ch^48); ch=getchar(); }return x*f;
}
const int N=2e5+7;
struct poi {ll x,y;poi (ll a=0,ll b=0) { x=a,y=b; }inline poi operator + (const poi &tmp) const { return poi(x+tmp.x,y+tmp.y); }inline poi operator - (const poi &tmp) const { return poi(x-tmp.x,y-tmp.y); }inline bool operator < (const poi &tmp) const { return x!=tmp.x ? x<tmp.x : y<tmp.y; }
}A[N],B[N],C[N],st[N],sum;
inline ll Cross(poi A,poi B) { return A.x*B.y-A.y*B.x; }
inline db Dot(poi A,poi B) { return A.x*B.x+A.y*B.y; }
inline db Len(poi A) { return sqrt(Dot(A,A)); }
inline bool cmp(const poi &A,const poi &B) { return Cross(A,B)>0||(Cross(A,B)==0&&Len(A)<Len(B)); }
void Tubao(poi *P,int &tot)
{sort(P+1,P+tot+1); sum=sum+P[1];/*如果此时求的是C的凸包P[1]=(0,0)没有贡献*/ for(int i=tot;i>=1;i--) P[i]=P[i]-P[1];sort(P+1,P+tot+1,cmp); int Top=0;for(int i=1;i<=tot;st[++Top]=P[i],i++)while(Top>1 && Cross(P[i]-st[Top-1],st[Top]-st[Top-1])>=0 ) Top--;tot=Top; for(int i=1;i<=tot;i++) P[i]=st[i];
}
int check(poi *P,int tot,poi A)
{A=A-sum;if(Cross(A,P[2])>0||Cross(P[tot],A)>0) return 0;int pos=lower_bound(P+1,P+tot+1,A,cmp)-P-1; if(pos==tot) return 0;return Cross(A-P[pos],P[pos+1]-P[pos])<=0;
}
int n,m,T,Q;
int main()
{n=read(),m=read(),Q=read();for(int i=1;i<=n;i++) A[i].x=read(),A[i].y=read();for(int i=1;i<=m;i++) B[i].x=-read(),B[i].y=-read();Tubao(A,n); Tubao(B,m);int la=1,lb=1; C[++T]=A[1]+B[1];while(la<=n||lb<=m){poi p1=A[la%n+1]+B[(lb-1)%m+1],p2=A[(la-1)%n+1]+B[lb%m+1];if(Cross(p1-C[T],p2-C[T])>=0) la++,C[++T]=p1;else lb++,C[++T]=p2;}Tubao(C,T);for(int i=1;i<=Q;i++){int x=read(),y=read();printf("%d\n",check(C,T,poi(x,y)));}return 0;
}