#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define se second
#define fi first
#include "robots.h"
int putaway(int a, int b, int t, int x[], int y[], int w[], int s[]) {
vector <pair<int,int>> vec;
for(int i=0;i<a;i++){
vec.pb({x[i],1});
}
for(int i=0;i<t;i++){
vec.pb({w[i],s[i]});
}
sort(vec.begin(),vec.end());
sort(y,y+b,greater<int>());
int l=0;
int r=t+1;
int ans=-1;
while(l<r){
int md=(l+r)>>1;
bool ok=1;
priority_queue<pair<int,int>>pq;
for(int i=0;i<(int)vec.size();i++){
if(vec[i].se==1){
for(int j=0;j<md;j++){
if(pq.size())pq.pop();
}
}
else{
pq.push(vec[i]);
}
}
priority_queue<int> pq2;
while(pq.size()){
pq2.push(pq.top().se);
pq.pop();
}
for(int i=0;i<b;i++){
if(!pq2.size())break;
if(pq2.top()>=y[i]){
ok=0;
break;
}
else{
for(int j=0;j<md;j++){
if(pq2.size())pq2.pop();
}
}
}
if(ok){
ans=md;
r=md;
}
else{
l=md+1;
}
}
return ans;
}
/*
---------------------------------------------------go get gold---------------------------------------------------------------------
- If u see the problem dp there's many option to iterate, u can iterate from 1- n or 1 - (possible max number)
- If u see the problem that has unexpected constraint u can divide it into 2 problem
- If u see the problem that can use binary search, then u can use binser + check condition
- in interactive problem there's many trick using binser
- kalo misal mau cari yang sama pake pernah ke visit ga bilangan itu
- kalo problem yang high itu biasanya optimisasi nya pake 2 array
- dp bisa aja kek kamu tenzing balls dimana optimisasi 2 dp
- janlup fibonacci
- kalo binser mending r = 3 * 1e18 aja
- kalo mau dibalik itu pake value nya tinggal diubah ke size
*/
