Problem: Suppose you have two polynomials <image> and <image>. Then polynomial <image> can be uniquely represented in the following way:

<image>

This can be done using [long division](https://en.wikipedia.org/wiki/Polynomial_long_division). Here, <image> denotes the degree of polynomial P(x). <image> is called the remainder of division of polynomial <image> by polynomial <image>, it is also denoted as <image>. 

Since there is a way to divide polynomials with remainder, we can define Euclid's algorithm of finding the greatest common divisor of two polynomials. The algorithm takes two polynomials <image>. If the polynomial <image> is zero, the result is <image>, otherwise the result is the value the algorithm returns for pair <image>. On each step the degree of the second argument decreases, so the algorithm works in finite number of steps. But how large that number could be? You are to answer this question. 

You are given an integer n. You have to build two polynomials with degrees not greater than n, such that their coefficients are integers not exceeding 1 by their absolute value, the leading coefficients (ones with the greatest power of x) are equal to one, and the described Euclid's algorithm performs exactly n steps finding their greatest common divisor. Moreover, the degree of the first polynomial should be greater than the degree of the second. By a step of the algorithm we mean the transition from pair <image> to pair <image>. 

Input

You are given a single integer n (1 ≤ n ≤ 150) — the number of steps of the algorithm you need to reach.

Output

Print two polynomials in the following format.

In the first line print a single integer m (0 ≤ m ≤ n) — the degree of the polynomial. 

In the second line print m + 1 integers between  - 1 and 1 — the coefficients of the polynomial, from constant to leading. 

The degree of the first polynomial should be greater than the degree of the second polynomial, the leading coefficients should be equal to 1. Euclid's algorithm should perform exactly n steps when called using these polynomials.

If there is no answer for the given n, print -1.

If there are multiple answer, print any of them.

Examples

Input

1


Output

1
0 1
0
1


Input

2


Output

2
-1 0 1
1
0 1

Note

In the second example you can print polynomials x2 - 1 and x. The sequence of transitions is

(x2 - 1, x) → (x, - 1) → ( - 1, 0).

There are two steps in it.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename... As>
struct tpl : public std::tuple<As...> {
  using std::tuple<As...>::tuple;
  tpl() {}
  tpl(std::tuple<As...> const& b) { std::tuple<As...>::operator=(b); }
  template <typename T = tuple<As...> >
  typename tuple_element<0, T>::type const& x() const {
    return get<0>(*this);
  }
  template <typename T = tuple<As...> >
  typename tuple_element<0, T>::type& x() {
    return get<0>(*this);
  }
  template <typename T = tuple<As...> >
  typename tuple_element<1, T>::type const& y() const {
    return get<1>(*this);
  }
  template <typename T = tuple<As...> >
  typename tuple_element<1, T>::type& y() {
    return get<1>(*this);
  }
  template <typename T = tuple<As...> >
  typename tuple_element<2, T>::type const& z() const {
    return get<2>(*this);
  }
  template <typename T = tuple<As...> >
  typename tuple_element<2, T>::type& z() {
    return get<2>(*this);
  }
  template <typename T = tuple<As...> >
  typename tuple_element<3, T>::type const& w() const {
    return get<3>(*this);
  }
  template <typename T = tuple<As...> >
  typename tuple_element<3, T>::type& w() {
    return get<3>(*this);
  }
};
using lli = long long int;
using llu = long long unsigned;
using pii = tpl<lli, lli>;
using piii = tpl<lli, lli, lli>;
using piiii = tpl<lli, lli, lli, lli>;
using vi = vector<lli>;
using vii = vector<pii>;
using viii = vector<piii>;
using vvi = vector<vi>;
using vvii = vector<vii>;
using vviii = vector<viii>;
template <class T>
using min_queue = priority_queue<T, vector<T>, greater<T> >;
template <class T>
using max_queue = priority_queue<T>;
template <size_t... I>
struct my_index_sequence {
  using type = my_index_sequence;
  static constexpr array<size_t, sizeof...(I)> value = {{I...}};
};
namespace my_index_sequence_detail {
template <typename I, typename J>
struct concat;
template <size_t... I, size_t... J>
struct concat<my_index_sequence<I...>, my_index_sequence<J...> >
    : my_index_sequence<I..., (sizeof...(I) + J)...> {};
template <size_t N>
struct make_index_sequence
    : concat<typename make_index_sequence<N / 2>::type,
             typename make_index_sequence<N - N / 2>::type>::type {};
template <>
struct make_index_sequence<0> : my_index_sequence<> {};
template <>
struct make_index_sequence<1> : my_index_sequence<0> {};
}  // namespace my_index_sequence_detail
template <class... A>
using my_index_sequence_for =
    typename my_index_sequence_detail::make_index_sequence<sizeof...(A)>::type;
template <class T, size_t... I>
void print_tuple(ostream& s, T const& a, my_index_sequence<I...>) {
  using swallow = int[];
  (void)swallow{0, (void(s << (I == 0 ? "" : ", ") << get<I>(a)), 0)...};
}
template <class T>
ostream& print_collection(ostream& s, T const& a);
template <class... A>
ostream& operator<<(ostream& s, tpl<A...> const& a);
template <class... A>
ostream& operator<<(ostream& s, tuple<A...> const& a);
template <class A, class B>
ostream& operator<<(ostream& s, pair<A, B> const& a);
template <class T, size_t I>
ostream& operator<<(ostream& s, array<T, I> const& a) {
  return print_collection(s, a);
}
template <class T>
ostream& operator<<(ostream& s, vector<T> const& a) {
  return print_collection(s, a);
}
template <class T, class U>
ostream& operator<<(ostream& s, multimap<T, U> const& a) {
  return print_collection(s, a);
}
template <class T>
ostream& operator<<(ostream& s, multiset<T> const& a) {
  return print_collection(s, a);
}
template <class T, class U>
ostream& operator<<(ostream& s, map<T, U> const& a) {
  return print_collection(s, a);
}
template <class T>
ostream& operator<<(ostream& s, set<T> const& a) {
  return print_collection(s, a);
}
template <class T>
ostream& print_collection(ostream& s, T const& a) {
  s << '[';
  for (auto it = begin(a); it != end(a); ++it) {
    s << *it;
    if (it != prev(end(a))) s << " ";
  }
  return s << ']';
}
template <class... A>
ostream& operator<<(ostream& s, tpl<A...> const& a) {
  s << '(';
  print_tuple(s, a, my_index_sequence_for<A...>{});
  return s << ')';
}
template <class... A>
ostream& operator<<(ostream& s, tuple<A...> const& a) {
  s << '(';
  print_tuple(s, a, my_index_sequence_for<A...>{});
  return s << ')';
}
template <class A, class B>
ostream& operator<<(ostream& s, pair<A, B> const& a) {
  return s << "(" << get<0>(a) << ", " << get<1>(a) << ")";
}
inline lli fexp(lli a, lli b, lli m) {
  lli r = 1;
  while (b) {
    if (b & 1) r = (r * a) % m;
    a = (a * a) % m;
    b /= 2;
  }
  return r;
}
inline lli invmod(lli a, lli m) { return fexp(a, m - 2, m); }
inline lli nmod(lli a, lli m) { return (a % m + m) % m; }
inline void nmode(lli& a, lli m) { a = nmod(a, m); }
vi poly_rem(vi A, vi B, lli m) {
  vi Q(0);
  vi R = A;
  while (!R.empty() && R.size() >= B.size()) {
    lli t = R.back() * invmod(B.back(), m) % m;
    while (Q.size() <= R.size() - B.size()) Q.push_back(0);
    Q[R.size() - B.size()] += t;
    while (!Q.empty() && Q.back() == 0) Q.pop_back();
    for (lli j = 0; j < (lli)(B.size()); ++j)
      nmode(R[j + R.size() - B.size()] -= t * B[j], m);
    while (!R.empty() && R.back() == 0) R.pop_back();
  }
  return R;
}
void poly_gcd(vi A, vi B, lli& step, lli m) {
  if (B.empty()) return;
  step++;
  poly_gcd(B, poly_rem(A, B, m), step, m);
}
lli primes[] = {1000000000 + 9};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  lli n;
  cin >> n;
  vi A(n + 1);
  A[n] = 1;
  vi B(n);
  for (lli k = 0; k < (lli)(1e7); ++k) {
    for (lli i = 0; i < (lli)(n); ++i) B[i] = rand() % 3 - 1;
    if (B.back() != 1) goto l1;
    for (lli k = 0; k < (lli)(1); ++k) {
      lli step = 0;
      poly_gcd(A, B, step, 1e9 + 7);
      if (step != n) goto l1;
    }
    {
      cout << A.size() - 1 << endl;
      for (lli x : A) cout << x << ' ';
      cout << endl;
      cout << B.size() - 1 << endl;
      for (lli x : B) cout << x << ' ';
      cout << endl;
      return 0;
    }
  l1:;
  }
  return 0;
}