#include "minerals.h"
// Knapsack DP is harder than FFT.
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii; typedef pair<ll,ll> pll;
#define ff first
#define ss second
#define pb emplace_back
#define FOR(i,n) for(int i = 0; i < (n); ++i)
#define FOO(i,a,b) for(int i = (a); i <= (b); ++i)
#define AI(x) begin(x),end(x)
template<class I> bool chmax(I &a, I b){ return a < b ? (a = b, true) : false;}
template<class I> bool chmin(I &a, I b){ return a > b ? (a = b, true) : false;}
#ifdef OWO
#define debug(args...) LKJ("[ " + string(#args) + " ]", args)
void LKJ(){ cerr<<endl;}
template<class I, class...T> void LKJ(I&&x, T&&...t){ cerr<<x<<", ", LKJ(t...);}
template<class I> void DE(I a, I b){ while(a < b) cerr<<*a<<" \n"[next(a) == b], ++a;}
#else
#define debug(...) 0
#define DE(...) 0
#endif
void solve(vector<int> &a, vector<int> &b){
debug("solving:");
DE(AI(a)); DE(AI(b));
if(a.size() < 1 or b.size() < 1){
return;
}
if(a.size() == 1 and b.size() == 1){
Answer(a[0], b[0]);
return;
}
vector<int> la, lb, ra, rb;
int h = a.size() / 2;
for(int i = 0; i < h; ++i) la.pb(a[i]);
for(int i = h; i < a.size(); ++i) lb.pb(a[i]);
int ls, cs;
for(int i: la) cs = Query(i);
assert(cs == la.size());
for(int i: b){
ls = cs;
cs = Query(i);
if(cs == ls) ra.pb(i);
else rb.pb(i);
}
for(int i: b)
cs = Query(i);
DE(AI(la)); DE(AI(lb));
DE(AI(ra)); DE(AI(rb));
for(int i: la) Query(i);
solve(la, ra);
solve(lb, rb);
}
void Solve(int N) {
vector<int> a, b;
int ls = 0, cs;
for(int i = 1; i <= N * 2; ++i){
cs = Query(i);
if(cs == ls) b.pb(i);
else a.pb(i);
ls = cs;
}
for(int i = 1; i <= N * 2; ++i)
Query(i);
solve(a, b);
}
