#include <bits/stdc++.h>
#include <string>
using namespace std;
using db = long double;
using ll = long long;
using pl = pair<ll, ll>;
using pi = pair<int, int>;
#define vt vector
#define f first
#define s second
#define pb push_back
#define all(x) x.begin(), x.end()
#define size(x) ((int) (x).size())
#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, b) FOR (i, 0, b)
#define endl '\n'
const ll INF = 1e18;
const int inf = 1e9;
template<typename... Args> // tuples
ostream& operator<<(ostream& os, tuple<Args...> t) {
apply([&](Args... args) { string dlm = "{"; ((os << dlm << args, dlm = ", "), ...); }, t);
return os << "}";
}
template<typename T, typename V> // pairs
ostream& operator<<(ostream& os, pair<T, V> p) { return os << "{" << p.f << ", " << p.s << "}"; }
template<class T, class = decltype(begin(declval<T>()))> // iterables
typename enable_if<!is_same<T, string>::value, ostream&>::type operator<<(ostream& os, const T& v) {
string dlm = "{";
for (auto& i : v) os << dlm << i, dlm = ", ";
return os << "}";
}
template <typename T, typename... V>
void printer(string pfx, const char *names, T&& head, V&& ...tail) {
int i = 0;
while (names[i] && names[i] != ',') i++;
constexpr bool is_str = is_same_v<decay_t<T>, const char*>;
if (is_str) cerr << " " << head;
else cerr << pfx, cerr.write(names, i) << " = " << head;
if constexpr (sizeof...(tail)) printer(is_str ? "" : ",", names + i + 1, tail...);
else cerr << endl;
}
#ifdef LOCAL
#define dbg(...) printer(to_string(__LINE__) + ": ", #__VA_ARGS__, __VA_ARGS__)
#else
#define dbg(x...)
#define cerr if (0) std::cerr
#endif
#include "fun.h"
/*
we can query distance and rooted subtree size
subtask 1 and 2: n^2
subtask 3: perfect binary tree, rooted at 0 with segtree indexing (kinda)
subtask 4: im pretty sure this is meant to be a star graph
repeatedly taking the diameter will give a valid tour trust
*/
int n, q;
vt<int> createFunTour(int N, int Q) {
n = N;
q = Q;
vt<vt<int>> dist(n, vt<int>(n));
F0R (i, n) {
FOR (j, i + 1, n) {
dist[i][j] = dist[j][i] = hoursRequired(i, j);
}
}
int u = 0, v = 0;
F0R (w, n) {
if (dist[u][w] > dist[u][v]) v = w;
}
swap(u, v);
// repeatedly find diameter, starting from u
vt<bool> done(n);
vt<int> out {u};
done[u] = true;
F0R (i, n - 1) {
v = u;
F0R (w, n) {
if (done[w]) continue;
if (dist[u][w] > dist[u][v]) v = w;
}
out.pb(v);
u = v;
done[v] = true;
}
dbg(out);
return out;
}
