#include <bits/stdc++.h>
#include "art.h"
#define fr(i, a, b) for (int i = (a); i <= (b); ++i)
#define rf(i, a, b) for (int i = (a); i >= (b); --i)
#define fe(x, y) for (auto& x : y)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define all(x) (x).begin(), (x).end()
#define sz(x) (int)(x).size()
#define pw(x) (1LL << (x))
using namespace std;
mt19937_64 rng(228);
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template <typename T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define fbo find_by_order
#define ook order_of_key
template <typename T>
bool umn(T& a, T b) {
return a > b ? a = b, 1 : 0;
}
template <typename T>
bool umx(T& a, T b) {
return a < b ? a = b, 1 : 0;
}
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template <typename T>
using ve = vector<T>;
void solve(int n) {
ve<int> pos(n + 1);
ve<int> order;
fr (i, 1, n) order.pb(i);
ve<int> vals;
fr (i, 1, n) {
vals.pb(publish(order));
if (i > 1) {
int val1 = vals.back();
int val2 = vals[sz(vals) - 2];
int d = val1 - val2;
int res = (n - 1 + d) / 2;
pos[order[0]] = res;
}
int x = order.back();
order.pop_back();
order.insert(order.begin(), x);
}
fill(all(order), 0);
fr (i, 2, n) {
order[pos[i]] = i;
}
fr (i, 0, sz(order) - 1) {
if (!order[i]) {
order[i] = 1;
break;
}
}
// fe (cur, order) cout << cur << " ";
// cout << "\n";
answer(order);
}
