Suppose you have two polynomials $A(x) = \sum_{k = 0}^{n} a_{k} x^{k}$ and $B(x) = \sum_{k = 0}^{m} b_{k} x^{k}$. Then polynomial $A(x)$ can be uniquely represented in the following way:$A(x) = B(x) \cdot D(x) + R(x), \operatorname{deg} R(x) < \operatorname{deg} B(x)$

This can be done using long division. Here, $\operatorname{deg} P(x)$ denotes the degree of polynomial P(x). $R(x)$ is called the remainder of division of polynomial $A(x)$ by polynomial $B(x)$, it is also denoted as $A \operatorname{mod} B$. 

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 $(A, B)$. If the polynomial $B(x)$ is zero, the result is $A(x)$, otherwise the result is the value the algorithm returns for pair $(B, A \operatorname{mod} B)$. 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 $(A, B)$ to pair $(B, A \operatorname{mod} B)$. 


-----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 x^2 - 1 and x. The sequence of transitions is(x^2 - 1, x) → (x,  - 1) → ( - 1, 0).

There are two steps in it.

Difficulty: competition

Starter code:

Solutions:
n = int(input())

a, b = [1], [0]

for i in range(n):
	new_b = a[:]
	a1 = a[:]
	a2 = a[:]
	a1.append(0)
	a2.append(0)
	for i in range(-1, -len(b) - 1, -1):
		a1[i] += b[i]
	for i in range(-1, -len(b) - 1, -1):
		a2[i] -= b[i]
	if max([abs(kek) for kek in a1]) < 2:
		a = a1
	elif max([abs(kek) for kek in a2]) < 2:
		a = a2
	else:
		print("oops")
		return
	b = new_b
print(len(a) - 1)
print(*(a[::-1]))
print(len(b) - 1)
print(*(b[::-1]))

# python3
# utf-8

n = int(input())
A = [1]
B = []
for _ in range(n):
    new_A = [0] + A[:]
    new_B = A[:]
    for i, b in enumerate(B):
        new_A[i] += b
    A = [a % 2 for a in new_A]
    B = [b % 2 for b in new_B]
print(len(A) - 1)
print(*A)
print(len(B) - 1)
print(*B)


def ii():
    return int(input())
def mi():
    return list(map(int, input().split()))
def li():
    return list(mi())

n = ii()
def gen(p, q, r):
    x = [0, p]
    y = [q]
    for c in range(1, n):
        
        nx = [0] + [-xi for xi in x]
        px = [0] + x
        if r: nx, px = px, nx
        
        bad = 0
        for i in range(len(y)):
            px[i] += y[i]
            if abs(px[i]) > 1:
                bad = 1
                break
        if not bad:
            x, y = px, x
            continue
        
        bad = 0
        for i in range(len(y)):
            nx[i] += y[i]
            if abs(nx[i]) > 1:
                bad = 1
                break
        if not bad:
            x, y = nx, x
            continue
        
        print('OH NO')
        z = 1 / 0
    return x, y

found = 0
for i, j in [(1, 1), (1, -1), (-1, 1), (-1, -1)]:
    for r in [0, 1]:
        x, y = gen(i, j, r)
        #print(x, y)
        if x[-1]==1 and y[-1]==1:
            found = 1
            break
    if found: break
if not found:
    print(n, 'OH NO!!')
print(len(x)-1)
print(*x)
print(len(y)-1)
print(*y)


n=int(input())
p = [[1], [0, 1]]
for i in range(n-1):
    t=[0]+p[-1]
    for j in range(len(p[i])):
        t[j]^=p[i][j]
    p.append(t)
print(n)
print(*p[n])
print(n-1)
print(*p[-2])


n = int(input())

def print_poly(a):
    print(len(a) - 1)
    print(' '.join(map(str, a)))

def shift_add(a, mul, b):
    c = [0] * (len(a) + 1)
    for i in range(len(a)): c[i + 1] = a[i] * mul
    for i in range(len(b)): c[i] += b[i]
    return c

a = [0, 1]
b = [1]

for i in range(n - 1):
    c = shift_add(a, 1, b)
    if max(c) <= 1 and min(c) >= -1:
        a, b = c, a
    else:
        c = shift_add(a, -1, b)
        if max(c) <= 1 and min(c) >= -1:
            a, b = c, a
        else:
            print('> <')

if a[-1] == -1:
    a = list([-x for x in a])
if b[-1] == -1:
    b = list([-x for x in b])

print_poly(a)
print_poly(b)



import sys

n = int(sys.stdin.readline().split()[0])

class Polynomial:
    def __init__(self, coef):
        first_nonzero = False
        index = len(coef) - 1
        while not first_nonzero:
            if not coef[index] == 0:
                first_nonzero = True
            else:
                if index == 0:
                    first_nonzero = True
                else:
                    index -= 1
        self.degree = index
        self.coef = [coef[j] for j in range(index + 1)]
    def multiply_by_x(self):
        new_coef = [0]
        for j in range(self.degree + 1):
            new_coef.append(self.coef[j])
        return Polynomial(new_coef)
    def minus(self):
        new_coef = [-self.coef[j] for j in range(self.degree + 1)]
        return Polynomial(new_coef)
    def add(self, other):
        other_coef = other.coef
        new_coef = [0 for j in range(max(self.degree, other.degree) + 1)]
        m = min(self.degree, other.degree)
        M = max(self.degree, other.degree)
        if self.degree > other.degree:
            bigger_poly = self
        else:
            bigger_poly = other
        for j in range(m + 1):
            new_coef[j] = self.coef[j] + other.coef[j]
        for j in range(m + 1, M+1):
            new_coef[j] = bigger_poly.coef[j]
            
        return Polynomial(new_coef) 
    def is_legal(self):
        result = True
        bools = [None for j in range(self.degree + 1)]
        bools[self.degree] = self.coef[self.degree] == 1
        for j in range(self.degree):
            bools[j] = self.coef[j] == 0 or self.coef[j] == 1 or self.coef[j] == -1
        for j in range(self.degree + 1):
            result = result and bools[j]
        return result
    def print(self):
        output = ""
        for j in range(self.degree + 1):
            output += str(self.coef[j]) + " "
        print(output)
            
        

f = []

f.append(Polynomial([1]))
f.append(Polynomial([0, 1]))

for j in range(2, 151):
    xf = f[j-1].multiply_by_x()
    t_1 = xf.add(f[j - 2])
    t_2 = xf.add(f[j - 2].minus())
    if t_1.is_legal():
        f.append(t_1)
    elif t_2.is_legal():
        f.append(t_2)
    #print(":(")


print(f[n].degree)
f[n].print()
print(f[n-1].degree)
f[n-1].print()

#for j in range(len(f)):
    #f[j].print()


f = [[1], [0, 1]]

n = int(input())

for i in range(2, n + 1):
	l = [0] + f[i - 1]
	for j in range(len(f[i - 2])):
		l[j] = (l[j] + f[i - 2][j]) & 1
	f.append(l)

print(n)
print(*f[n])
print(n - 1)
print(*f[n - 1])



# Made By Mostafa_Khaled