#include "souvenirs.h"
#include <bits/stdc++.h>
using namespace std;
#define MOD1 1000000007
#define MOD2 998244353
#define fir first
#define sec second
#define pushf push_front
#define pushb push_back
#define popf pop_front
#define popb pop_back
#define mp make_pair
#define all(a) a.begin(), a.end()
#define FOR1(a) for (int _ = 0; _ < a; ++_)
#define FOR2(i, a) for (int i = 0; i < a; ++i)
#define FOR3(i, a, b) for (int i = a; i < b; ++i)
#define RFOR1(a) for (int _ = (a)-1; _ >= 0; --_)
#define RFOR2(i, a) for (int i = (a)-1; i >= 0; --i)
#define RFOR3(i, a, b) for (int i = (b)-1; i >= a; --i)
#define overload3(a, b, c, d, ...) d
// Always choose the fourth argument to call. Hence, which function to call is determined by the number of given arguments.
#define REP(...) overload3(__VA_ARGS__, FOR3, FOR2, FOR1)(__VA_ARGS__)
#define RREP(...) overload3(__VA_ARGS__, RFOR3, RFOR2, RFOR1)(__VA_ARGS__)
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef pair<ll, ll> pll;
typedef vector<ll> vll;
typedef vector<bool> vb;
typedef vector<char> vc;
typedef vector<string> vs;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef vector<vi> vvi;
typedef vector<vll> vvll;
typedef vector<vb> vvb;
typedef vector<vc> vvc;
typedef vector<vpii> vvpii;
typedef vector<vpll> vvpll;
typedef queue<int> qi;
typedef queue<ll> qll;
typedef queue<pii> qpii;
typedef queue<pll> qpll;
typedef deque<int> dqi;
typedef deque<ll> dqll;
typedef deque<pii> dqpii;
typedef deque<pll> dqpll;
typedef priority_queue<int> pqi;
typedef priority_queue<ll> pqll;
typedef priority_queue<pii> pqpii;
typedef priority_queue<pll> pqpll;
typedef priority_queue<int, vi, greater<int> > r_pqi;
typedef priority_queue<ll, vll, greater<ll> > r_pqll;
typedef priority_queue<pii, vpii, greater<pii> > r_pqpii;
typedef priority_queue<pll, vpll, greater<pll> > r_pqpll;
int n;
vll cost;
vi cnt;
int solve(ll val, int now) {
auto [res, rem] = transaction(val);
REP(i, res.size()) {
cnt[res[i]]++;
}
ll used = val - rem;
while(res.back() >= now){
used -= cost[res.back()];
res.popb();
}
while(res.front() + 1 < now){
now = solve((used - 1) / res.size(), now);
while(res.back() >= now){
used -= cost[res.back()];
res.popb();
}
}
cost[res.front()] = used;
return res.front();
}
void buy_souvenirs(int N, ll P0) {
n = N;
cost = vll(n, -1);
cost[0] = P0;
cnt = vi(n, 0);
solve(P0-1, N);
REP(i, n) {
REP(j, cnt[i], i) {
transaction(cost[i]);
}
}
}
