#include "monster.h"
#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define MOD (1000000000+7)
#define MOD1 (998244353)
#define pb push_back
#define all(x) x.begin(), x.end()
#define en cout << '\n'
#define ff first
#define ss second
#define pii pair<int,int>
#define vi vector<int>
const int N = 1e6+100, M = 1e5+10, K = 52, MX = 30;
namespace {
void rev(vector<int> &V, int l, int r){
for(int i = l; i <= (l+r)/2; ++i){
swap(V[i], V[r - (i-l)]);
}
}
} // namespace
vector<int> Solve(int n) {
// vi v(n), res(n);
// vector<vi> RES(n);
// for(int i = 0; i < n; ++i){
// for(int j = i + 1; j < n; ++j){
// bool x = Query(i, j);
// x ? v[i]++ : v[j]++;
// }
// }
// for(int i = 0; i < n; ++i){
// if(v[i] != 1 && v[i] != n-2){
// res[i] = v[i];
// RES[v[i]].pb(i);
// }else{
// RES[v[i]].pb(i);
// }
// }
// bool x = !Query(RES[1][0], RES[2][0]) && (RES[2].size() == 1 ? true : !Query(RES[1][0], RES[2][1])); // whether it's beaten by both
// if(x){
// res[RES[1][0]] = 0;
// res[RES[1][1]] = 1;
// }else{
// res[RES[1][0]] = 1;
// res[RES[1][1]] = 0;
// }
// bool y = Query(RES[n-2][0], RES[n-3][0]) && (RES[n-3].size() == 1 ? true : Query(RES[n-2][0], RES[n-3][1])); // whether it's beats both
// if(y){
// res[RES[n-2][0]] = n-1;
// res[RES[n-2][1]] = n-2;
// }else{
// res[RES[n-2][0]] = n-2;
// res[RES[n-2][1]] = n-1;
// }
vi res(n, -1);
vector<int> V;
V.pb(0);
for(int i = 1; i < n; ++i){
int l = 0, r = int(V.size())-1, pos = V.size();
while(l <= r){
int mid = l+r>>1;
if(!Query(i, V[mid])){
pos = mid;
r = mid - 1;
}else l = mid + 1;
}
V.insert(V.begin() + pos, i);
}
// now we got chains...
// what do we do.. we got around N~ queries.
// identify the first chain
int last2 = 0, beat = 0;
for(int i = 1; i < n; ++i){
if(Query(V[0], V[i])){
++beat;
last2 = i;
}
}
if(beat == 1){
cerr << "Case 1\n";
// hardest case ngl
if(last2 == 1){
cerr << "Case 1.1\n";
res[V[0]] = 0;
res[V[1]] = 1;
int largest = V[1], last = 1, j = 2;
int cur = 2;
for(; j < n; ++j){
if(Query(largest, V[j])){
// this is cur
res[V[j]] = cur++;
for(int l = j - 1; l > last; --l){
res[V[l]] = cur++;
}
largest = V[last + 1];
last = j;
}
}
}else if(last2 > 2){
cerr << "Case 1.2\n";
bool bb = Query(V[1], V[last2]);
if(bb){
// 0 k k-1 ... 1
res[V[0]] = 0;
res[V[last2]] = 1;
int cur = 2;
for(int j = last2 - 1; j > 0; --j){
res[V[j]] = cur++;
}
int largest = V[1], last = last2, j = last2 + 1;
for(; j < n; ++j){
if(Query(largest, V[j])){
// this is cur
res[V[j]] = cur++;
for(int l = j - 1; l > last; --l){
res[V[l]] = cur++;
}
largest = V[last + 1];
last = j;
}
}
}else{
cerr << "Case 1.3\n";
res[V[0]] = 1;
res[V[1]] = 0;
res[V[last2]] = 2;
int cur = 3;
for(int j = last2 - 1; j > 1; --j){
res[V[j]] = cur++;
}
int largest = V[2], last = last2, j = last2 + 1;
for(; j < n; ++j){
if(Query(largest, V[j])){
// this is cur
res[V[j]] = cur++;
for(int l = j - 1; l > last; --l){
res[V[l]] = cur++;
}
largest = V[last + 1];
last = j;
}
}
}
}else{
cerr << "jump from the window\n";
// idk jump from the window
// 0 2 1 or 1 0 2, all 3 queries return the same...
int beat2 = 0;
for(int i = 0; i < n; ++i) if(i != 1 && Query(V[1], V[i])) ++beat2;
int cur = 3, last = 2, largest, j = 3;
if(beat2 == 1){
// 1 0 2
res[V[0]] = 1;
res[V[1]] = 0;
res[V[2]] = 2;
largest = V[2];
}else{
res[V[0]] = 0;
res[V[1]] = 2;
res[V[2]] = 1;
largest = V[1];
}
for(; j < n; ++j){
if(Query(largest, V[j])){
// this is cur
res[V[j]] = cur++;
for(int l = j - 1; l > last; --l){
res[V[l]] = cur++;
}
largest = V[last + 1];
last = j;
}
}
}
}else if(beat == n-2){
cerr << "Case 2\n";
bool bb = Query(V.back(), V[1]);
if(bb){
// n-2 ... 0 n-1 case
int cur = 0;
for(int i = n-2; i >= 0; --i) res[V[i]] = cur++;
res[V.back()] = n-1;
}else{
// n-1 n-2... 0 case
int cur = 0;
for(int i = n-1; i >= 0; --i) res[V[i]] = cur++;
}
}else{
cerr << "Case 3\n";
res[V[0]] = beat;
int cur = beat - 1;
int j = 1;
for(; cur >= 0; ++j){
res[V[j]] = cur--;
}
int largest = V[0], last = j - 1;
cur = res[V[0]] + 1;
for(; j < n; ++j){
if(Query(largest, V[j])){
// this is cur
res[V[j]] = cur++;
for(int l = j - 1; l > last; --l){
res[V[l]] = cur++;
}
largest = V[last + 1];
last = j;
}
}
}
return res;
}
