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;
using ll = std::int_fast64_t;
using P = pair<ll, ll>;
constexpr ll inf = 1000000000;
constexpr ll mod = 1000000007;
constexpr long double eps = 1e-15;
template <typename T1, typename T2>
ostream& operator<<(ostream& os, pair<T1, T2> p) {
  os << to_string(p.first) << " " << to_string(p.second);
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, vector<T>& v) {
  for (ll i = 0; i < (int)v.size(); i++) {
    if (i) os << " ";
    os << to_string(v[i]);
  }
  return os;
}
struct modint {
  ll n;

 public:
  modint(const ll n = 0) : n((n % mod + mod) % mod) {}
  static modint pow(modint a, int m) {
    modint r = 1;
    while (m > 0) {
      if (m & 1) {
        r *= a;
      }
      a = (a * a);
      m /= 2;
    }
    return r;
  }
  modint& operator++() {
    *this += 1;
    return *this;
  }
  modint& operator--() {
    *this -= 1;
    return *this;
  }
  modint operator++(int) {
    modint ret = *this;
    *this += 1;
    return ret;
  }
  modint operator--(int) {
    modint ret = *this;
    *this -= 1;
    return ret;
  }
  modint operator~() const { return (this->pow(n, mod - 2)); }
  friend bool operator==(const modint& lhs, const modint& rhs) {
    return lhs.n == rhs.n;
  }
  friend bool operator<(const modint& lhs, const modint& rhs) {
    return lhs.n < rhs.n;
  }
  friend bool operator>(const modint& lhs, const modint& rhs) {
    return lhs.n > rhs.n;
  }
  friend modint& operator+=(modint& lhs, const modint& rhs) {
    lhs.n += rhs.n;
    if (lhs.n >= mod) lhs.n -= mod;
    return lhs;
  }
  friend modint& operator-=(modint& lhs, const modint& rhs) {
    lhs.n -= rhs.n;
    if (lhs.n < 0) lhs.n += mod;
    return lhs;
  }
  friend modint& operator*=(modint& lhs, const modint& rhs) {
    lhs.n = (lhs.n * rhs.n) % mod;
    return lhs;
  }
  friend modint& operator/=(modint& lhs, const modint& rhs) {
    lhs.n = (lhs.n * (~rhs).n) % mod;
    return lhs;
  }
  friend modint operator+(const modint& lhs, const modint& rhs) {
    return modint(lhs.n + rhs.n);
  }
  friend modint operator-(const modint& lhs, const modint& rhs) {
    return modint(lhs.n - rhs.n);
  }
  friend modint operator*(const modint& lhs, const modint& rhs) {
    return modint(lhs.n * rhs.n);
  }
  friend modint operator/(const modint& lhs, const modint& rhs) {
    return modint(lhs.n * (~rhs).n);
  }
};
istream& operator>>(istream& is, modint m) {
  is >> m.n;
  return is;
}
ostream& operator<<(ostream& os, modint m) {
  os << m.n;
  return os;
}
long long extgcd(long long a, long long b, long long& x, long long& y) {
  long long d = a;
  if (b != 0) {
    d = extgcd(b, a % b, y, x);
    y -= (a / b) * x;
  } else {
    x = 1;
    y = 0;
  }
  return d;
}
long long mod_inverse(long long a, long long m) {
  long long x, y;
  if (extgcd(a, m, x, y) == 1)
    return (m + x % m) % m;
  else
    return -1;
}
vector<long long> fact(3030303 + 1, inf);
long long mod_fact(long long n, long long& e) {
  if (fact[0] == inf) {
    fact[0] = 1;
    if (3030303 != 0) fact[1] = 1;
    for (ll i = 2; i <= 3030303; ++i) {
      fact[i] = (fact[i - 1] * i) % mod;
    }
  }
  e = 0;
  if (n == 0) return 1;
  long long res = mod_fact(n / mod, e);
  e += n / mod;
  if ((n / mod) % 2 != 0) return (res * (mod - fact[n % mod])) % mod;
  return (res * fact[n % mod]) % mod;
}
long long mod_comb(long long n, long long k) {
  if (n < 0 || k < 0 || n < k) return 0;
  long long e1, e2, e3;
  long long a1 = mod_fact(n, e1), a2 = mod_fact(k, e2),
            a3 = mod_fact(n - k, e3);
  if (e1 > e2 + e3) return 0;
  return (a1 * mod_inverse((a2 * a3) % mod, mod)) % mod;
}
using mi = modint;
mi mod_pow(mi a, ll n) {
  mi ret = 1;
  mi tmp = a;
  while (n > 0) {
    if (n % 2) ret *= tmp;
    tmp = tmp * tmp;
    n /= 2;
  }
  return ret;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  deque<int> dq1, dq2;
  dq1.push_back(1);
  dq2.push_back(0);
  for (ll i = 0; i < (int)n; i++) {
    deque<int> dq3 = dq1;
    dq1.push_front(0);
    bool b = false;
    for (ll j = 0; j < (int)dq2.size(); j++) {
      if (dq1[j] + dq2[j] > 1 || dq1[j] + dq2[j] < -1) b = true;
    }
    for (ll j = 0; j < (int)dq2.size(); j++) {
      if (!b)
        dq1[j] += dq2[j];
      else
        dq1[j] -= dq2[j];
    }
    dq2 = dq3;
  }
  cout << (int)dq1.size() - 1 << endl;
  for (ll i = 0; i < (int)dq1.size(); i++) {
    if (i) cout << " ";
    cout << dq1[i];
  }
  cout << endl;
  cout << (int)dq2.size() - 1 << endl;
  for (ll i = 0; i < (int)dq2.size(); i++) {
    if (i) cout << " ";
    cout << dq2[i];
  }
  cout << endl;
  return 0;
}