#include "robots.h"
/*
Author of all code: Pedro BIGMAN Dias
Last edit: 15/02/2021
*/
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#pragma GCC optimize("Ofast")
#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <string>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <queue>
#include <deque>
#include <list>
#include <iomanip>
#include <stdlib.h>
#include <time.h>
#include <cstring>
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ld;
#define REP(i,a,b) for(ll i=(ll) a; i<(ll) b; i++)
#define pb push_back
#define mp make_pair
#define pl pair<ll,ll>
#define ff first
#define ss second
#define whole(x) x.begin(),x.end()
#define DEBUG(i) cout<<"Pedro Is The Master "<<i<<endl
#define INF 500000000LL
#define EPS 0.00000001
#define pi 3.14159
ll mod=1000000007LL;
template<class A=ll>
void Out(vector<A> a) {REP(i,0,a.size()) {cout<<a[i]<<" ";} cout<<endl;}
template<class A=ll>
void In(vector<A> &a, ll N) {A cur; REP(i,0,N) {cin>>cur; a.pb(cur);}}
vector<vector<ll> > ps;
ll Query(ll x1,ll x2, ll y1, ll y2)
{
ll ans = ps[x2][y2];
if(y1>0) {ans-=ps[x2][y1-1];}
if(x1>0) {ans-=ps[x1-1][y2];}
if(x1>0 && y1>0) {ans+=ps[x1-1][y1-1];}
return ans;
}
int putaway(int a, int b, int T, int X[], int Y[], int W[], int S[])
{
vector<ll> A,B; vector<pl> p;
REP(i,0,a) {A.pb((ll) X[i]);} REP(i,0,b) {B.pb((ll) Y[i]);}
REP(i,0,T) {p.pb((pl) {(ll) W[i], (ll) S[i]});}
A.pb(0LL); B.pb(0LL); sort(whole(A)); sort(whole(B));
vector<vector<ll> > oc; vector<ll> xx; REP(i,0,B.size()) {xx.pb(0);} REP(i,0,A.size()) {oc.pb(xx);}
ll indA,indB;
REP(i,0,T)
{
indA = (ll) (upper_bound(whole(A),p[i].ff) - A.begin()) -1LL;
indB = (ll) (upper_bound(whole(B),p[i].ss) - B.begin()) -1LL;
oc[indA][indB]++;
}
REP(i,0,A.size()) {ps.pb(xx);}
REP(i,0,A.size())
{
REP(j,0,B.size())
{
if(i==0 && j==0) {ps[i][j]=oc[i][j];}
else if(i==0) {ps[i][j]=ps[i][j-1]+oc[i][j];}
else if(j==0) {ps[i][j]=ps[i-1][j]+oc[i][j];}
else {ps[i][j]=ps[i-1][j]+ps[i][j-1]-ps[i-1][j-1]+oc[i][j];}
}
}
if(oc[A.size()-1][B.size()-1]!=0) {return -1;}
ll ans=0LL;
ll val; ll sizeA,sizeB;
REP(i,0,A.size())
{
REP(j,0,B.size())
{
if(i==A.size()-1 && j==B.size()-1) {continue;}
sizeA = A.size()-i; sizeB = B.size()-j;
val = (Query(i,A.size()-1,j,B.size()-1)+sizeA+sizeB-3LL)/(sizeA+sizeB-2LL);
ans=max(ans,val);
}
}
return ans;
}
