#pragma GCC optimize(3,"Ofast","inline")
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
const double EPS=1e-7;
const double PI=acos(-1);
const ll MOD=1000000007;
int _IO=[](){ios::sync_with_stdio(0);cin.tie(0); cout.tie(0);return 0;}();
#define REP(i, n) for (ll i = 0; i < n; i++)
#define REPR(i, n) for (ll i = n; i >= 0; i--)
#define FOR(i, m, n) for (ll i = m; i < n; i++)
#define FORR(i, m, n) for (ll i = m; i >= n; i--)
#define ALL(v) v.begin(), v.end()
#define ALLR(v) v.rbegin(), v.rend()
#define pb push_back
template <class T>bool chmax(T& a, const T& b){if (a < b){a = b;return 1;}else return 0;}
template <class T>bool chmin(T& a, const T& b){if (b < a){a = b;return 1;}else return 0;}
//判断从cur到next再到query的拐弯方向
//左拐返回2，右拐返回4，不变线的话前进返回1，后退小步返回-1，后退大步返回3
template<class T>
int getDirection(T*cur,T*next,T*query){
    T x1=*next-*cur;
    T y1=*(next+1)-*(cur+1);
    T x2=*query-*next;
    T y2=*(query+1)-*(next+1);
    T res=x1*y2-x2*y1;
    if(res>0)return 2;
    else if(res<0)return 4;
    res=x1*x2+y1*y2;
    if(res>0)return 1;
    else if(abs(x2)<=abs(x1)&&abs(y2)<=abs(y1))return -1;
    else return 3;
}


const int N=500;
int n,m;
pair<int,int>A[N],B[N];
int mat[N][N];
int solve(){
    memset(mat,0x3f,sizeof(mat));
    REP(i,m)REP(j,i){
        bool l=false,r=false,out=false;
        REP(k,n){
            int a=getDirection(&B[i].first,&B[j].first,&A[k].first);
            if(a==2){l=true;if(r)break;}
            else if(a==4){r=true;if(l)break;}
            else if(a!=-1){out=true;break;}
        }
        if(l&&r)continue;
        if(out)continue;
        if(!l&&!r)mat[i][j]=mat[j][i]=1;
        else if(l)mat[i][j]=1;
        else mat[j][i]=1;
    }
    int ans=0x3f3f3f3f;
    REP(k,m)
        REP(i,m)if(mat[i][k]<0x3f3f3f3f)
            REP(j,m)
                chmin(mat[i][j],mat[i][k]+mat[k][j]);
    REP(k,m)
        chmin(ans,mat[k][k]);
    return ans;
}

int main(){
    while(~scanf("%d",&n)){
        REP(i,n)scanf("%d %d",&A[i].first,&A[i].second);
        scanf("%d",&m);
        REP(i,m)scanf("%d %d",&B[i].first,&B[i].second);
        int ans=solve();
        if(ans>m)printf("ToT\n");
        else printf("%d\n",m-ans);
    }
}