#include "ricehub.h"
//#include "grader.cpp"
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimize("Ofast,unroll-loops")
using namespace std;
using namespace __gnu_pbds;
#define pb push_back
#define all(x) x.begin(),x.end()
#define ar array
#define mrand(a, b) uniform_int_distribution<int>(a, b)(rng)
template<class T>bool umax(T &a,T b){if(a<b){a=b;return true;}return false;}
template<class T>bool umin(T &a,T b){if(b<a){a=b;return true;}return false;}
template<class T> using ste = tree<T, null_type, less_equal<T>,
rb_tree_tag, tree_order_statistics_node_update>;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
namespace FAST {
template<typename T, typename F>
istream &operator>>(istream &cin, pair<T, F> &p) {
cin >> p.first >> p.second;
return cin;
}
template<typename T, typename F>
ostream &operator<<(ostream &cout, pair<T, F> &p) {
cout << p.first << ' ' << p.second;
return cout;
}
template<typename T>
istream &operator>>(istream &cin, vector<T> &a) {
for (T &i: a) cin >> i;
return cin;
}
template<typename T>
ostream &operator<<(ostream &cout, vector<T> &a) {
for (T i: a) cout << i << ' ';
return cout;
}
template<typename T>
istream &operator>>(istream &cin, deque<T> &a) {
for (T &i: a) cin >> i;
return cin;
}
template<typename T>
ostream &operator<<(ostream &cout, deque<T> &a) {
for (T i: a) cout << i << ' ';
return cout;
}
}
using namespace FAST;
const long long inf = 1e17 + 7;
const int mod = 1e9 + 7;
const int md = 998244353;
int besthub(int R, int L, int X[], long long B)
{
vector<long long>pref(R+1), suff(R+2);
for(int i = 0;i<R;i++){
pref[i] = ((i == 0) ? 0 : pref[i-1]);
pref[i] += ((i == 0) ? 0 : (X[i] - X[i-1]) * i);
}
for(int i = R-1;i>=0;i--){
suff[i] = suff[i+1];
suff[i] += ((i == R-1) ? 0 : (X[i+1] - X[i]) * (R - i - 1));
}
int res = 1;
for(int i = 0;i<R;i++){
int l = 0, r = i-1, ans = -1;
while(l <= r){
long long mid = (l + r) / 2;
long long sum = 0, u;
u = i - (i-mid) / 2;
sum = pref[u] + suff[u];
if(mid > 0){
sum -= 0ll + pref[mid] + mid * (X[u] - X[mid]);
}
if(mid < R - 1){
sum -= 0ll + suff[i] + (R - i - 1) * (X[i] - X[u]);
}
// if(i == R - 1 && mid == R - 3)cout << sum << "\n";
if(sum <= B){
r = mid - 1;
ans = mid;
}
else l = mid + 1;
}
if(ans != -1){
umax(res, (i - ans + 1));
}
}
return res;
}
