#pragma GCC optimize("Ofast")
#pragma GCC optimize("O3,unroll-loops")
#include <bits/stdc++.h>
using namespace std;
#define f0r(i, n) for (auto i = 0; i < (n); ++i)
#define fnr(i, n, k) for (auto i = (n); i < (k); ++i)
#define all(v) (v).begin(), (v).end()
#define pb push_back
#define F first
#define S second
#define ctn(x) cout << x << '\n'
#define forl(a, l) for (auto a : l)
#define ctl(l) for (auto &a : (l)) cout << a << ' '; cout << endl;
#define lb(v, x) (lower_bound(all(v), x) - begin(v))
#define ub(v, x) (upper_bound(all(v), x) - begin(v))
#define pq priority_queue
template <class T>
using V = vector<T>;
using ll = long long;
using vi = V<int>;
using vl = V<ll>;
using pi = pair<int, int>;
using ti = tuple<int, int, int>;
using Adj = V<vi>;
using vvi = V<vi>;
#include "doll.h"
void create_circuit(int M, vi A) {
vi C(M + 1);
A.pb(0);
vi X, Y;
int s = 0;
map<int, vi> mp;
int p = 0;
forl(x, A) {
mp[p].pb(x);
p = x;
}
auto F = [&](int a, int b) {
X.pb(a);
Y.pb(b);
return -X.size();
};
for (auto &[x, v]: mp) {
if (v.size() == 1) C[x] = v[0];
else if (v.size() == 2) {
C[x] = F(v[0], v[1]);
}
else {
int n = v.size(), m = 1;
while (m < n) m *= 2;
//cout << "CHEK: " << n << ' ' << m << endl;
s = X.size()-1;
X.resize(X.size()+m-1);
Y.resize(Y.size()+m-1);
fnr(i, 1, m/2) {
X[s+i] = -(s+(2*i) + 1);
Y[s+i] = -(s+(2*i+1) + 1);
}
fnr(i, m/2, m) {
//cout << "YO" << endl;
int j = (i-(m/2));
int a = j | (j+1);
if (j >= m/4) {
j -= m/4;
a = (j | (j+1))+1;
}
int b = a + m/2;
--a, --b;
X[s+i] = v[a];
if (b >= n-1) {
Y[s+i] = i == m-1 ? v[n-1] : -(s+1 + 1);
} else Y[s+i] = v[b];
}
C[x] = -(s + 1 + 1);
}
//else if (v.size() == 2) {
// C[x] = F(v[0], v[1]);
//} else if (v.size() == 3) {
// int t = -(s+3);
// C[x] = F(F(v[0], t), F(v[1], v[2]));
//} else {
// C[x] = F(F(v[0], v[2]), F(v[1], v[3]));
// continue;
// X.pb(-(s+2));
// X.pb(-(s+3));
// C[x] = -++s;
// X.pb(v[0]);
// Y.pb(v[2]);
// X.pb(v[1]);
// Y.pb(v[3]);
// s += 2;
//}
}
//ctl(C);
//ctl(X);
//ctl(Y);
answer(C, X, Y);
return;
//for (int i = 1; i <= M; ++i) {
// C[i] = 1;
//}
//for (int k = 0; k < N; ++k) {
// X[k] = Y[k] = A[k];
//}
answer(C, X, Y);
}
