// essa sol usa N + 2*log queries, pra ajeitar é so reverter a arvore da esquerda mas é casework chato q eu faco dps
#include "doll.h"
#include <algorithm>
#include <cstdio>
#include <cassert>
int x[400010], y[400010], cnt = 1;
void solve(int id, int pot, std::vector<int> A, bool esp) {
int N = (int)(A.size()) + esp;
if(!pot) {
if(N == 1 && esp) return (void)(x[id] = -1, y[id] = 0);
if(N == 2 && esp) A.push_back(0);
if(N == 1) return (void)(x[id] = A[0], y[id] = -1);
if(N == 2) return (void)(x[id] = A[0], y[id] = A[1]);
assert(0);
}
std::vector<int> esq, dir;
if(N-esp > (1 << pot)) {
y[id] = -(++cnt);
int aq = N - (1 << pot) - esp;
for(int i = 0; i < (int)(A.size()); i++)
if((i&1) && aq) --aq, esq.push_back(A[i]);
else dir.push_back(A[i]);
solve(-y[id], pot-1, esq, esp);
} else if(esp) {
x[id] = -1;
y[id] = -(++cnt);
solve(-y[id], pot-1, std::vector<int>(0), esp);
dir = A;
} else y[id] = -1, dir = A;
if(!dir.size()) return;
x[id] = -(++cnt);
solve(-x[id], pot-1, dir, 0);
}
void create_circuit(int M, std::vector<int> A) {
int N = (int)(A.size());
if(N == 1) return answer({1, 0}, {}, {});
std::vector<int> C(M + 1);
C[0] = A[0]; // A.push_back(0);
for (int i = 1; i <= M; ++i)
C[i] = -1;
std::reverse(A.begin(), A.end());
A.pop_back();
std::reverse(A.begin(), A.end());
solve(1, 31-__builtin_clz(N)-(__builtin_popcount(N) == 1), A, 1);
std::vector<int> X(cnt), Y(cnt);
for(int i = 1; i <= cnt; i++)
X[i-1] = x[i], Y[i-1] = y[i];
answer(C, X, Y);
}
