#include <bits/stdc++.h>
//#include "includeall.h"
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define all(x) x.begin(), x.end()
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define FOR(i, x, n) for (ll i =x; i<=n; ++i)
#define RFOR(i, x, n) for (ll i =x; i>=n; --i)
#pragma GCC optimize("O3","unroll-loops")
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
typedef long long ll;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<pi, ll> pii;
typedef pair<pi, pi> piii;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
ll n;
vector<ll> order;
int Query(const std::vector<int>& M);
ll query(ll mid, ll x)
{
vector<int> now(n, 0);
for (ll i=0; i<=mid; ++i) now[order[i]] = 1;
ll ans = Query(now);
now[x] = 1;
return Query(now)==ans;
}
void Answer(const std::vector<int>& res);
void Solve(int N)
{
vector<int> res(N, 1);
n = N;
if(n==1) {Answer(res); return;}
order.clear();
for (ll i=0; i<n; ++i) order.pb(i);
vector<int> now(n, 1);
shuffle(all(order), rng);
ll root = -1;
for (ll i=0; i<n; ++i)
{
now[order[i]] = 0;
if (Query(now)==1) {root = order[i]; break;}
now[order[i]] = 1;
}
for (ll i=0; i<n; ++i)
{
res[i] = root+1;
order.erase(find(all(order), root));
if (i==n-1) break;
show(root);
// find first
ll lo = 0, hi = order.size()-1;
while (lo!=hi)
{
ll mid = (lo + hi) >> 1;
if (query(mid, root)) hi = mid;
else lo = mid + 1;
}
if (lo < order.size()) root = order[lo];
}
Answer(res);
}
