// Difficult(y) Choice
#include <bits/stdc++.h>
#include "books.h"
using namespace std;
using ll=long long;
#define FOR(i, a, b) for(int i=a; i<b; i++)
#define ROF(i, a, b) for(int i=b-1; i>=a; i--)
#define F0R(i, n) FOR(i, 0, n)
#define R0F(i, n) ROF(i, 0, n)
#define f first
#define s second
#define pb push_back
#define siz(x) (int)x.size()
const int mxN=1e5+10;
int n, k, s; ll a;
ll x[mxN];
ll qry(int V){
if(x[V]!=-1) return x[V];
x[V]=skim(V+1);
return x[V];
}
void chk(int X){
if(X<0) return;
if(X+k>n) return;
ll sum=0;
vector<int> Ans;
FOR(i, X, X+k){
sum+=qry(i);
Ans.pb(i+1);
}
if(sum>=a&&sum<=2*a) {
answer(Ans);
exit(0);
}
}
void solve(int N, int K, ll A, int S){
n=N, k=K, a=A, s=S;
memset(x, -1, sizeof x);
{
ll sum=0;
F0R(i, k-1) sum+=qry(i);
int lo=k-1, hi=n;
while(lo<hi){
int m=(lo+hi)/2;
// cerr << m << endl;
if(qry(m)+sum>2*a) hi=m-1;
else if(qry(m)+sum<a) lo=m+1;
else{
vector<int> Ans; F0R(i, k-1) Ans.pb(i+1);
Ans.pb(m+1);
answer(Ans); exit(0);
}
}
n=hi;
}
int lo=0, hi=n;
while(lo+1<hi){
int m=(lo+hi)/2;
if(qry(m)*k>=a) hi=m;
else lo=m+1;
}
FOR(i, lo-k+1, lo+1){
chk(i);
}
impossible();
// assert(n==s);
// task is query 40 books and try to see if we can get a set of K books such that A<=\sum<=2A we have K is less than 10 so SMALL
// solve subtask by subtask:
/*
observation, problem has to require BINARY SEARCH
subtask 1: we can just like iterte over first 2 books, check which if we have a third one
subtask 2: if the last element we take is <=A then if the previous elements are >A, we can find a solution
current solution:
binary search for first X such that x[X]*K>=A then notice this element has to be used
at least I think that is true
20+17 queries minus a couple
thing is we also need to check for the smallest number with value >= A
We probably have 80 points
*/
// vector<ll> x(n);
// F0R(i, n){
// x[i]=skim(i+1);
// }
// ll sum=0;
// F0R(i, k-1) sum+=qry(i);
// FOR(i, k-1, n){
// sum+=qry(i);
// if(sum>=a&&sum<=2*a){
// vector<int> ans;
// FOR(j, i-k+1, i+1) ans.pb(j+1);
// answer(ans);
// }
// sum-=qry(i-k+1);
// }
// impossible();
}
// int main(){
// cin.tie(0)->sync_with_stdio(0);
// }
