# ポラード・ロー素因数分解法 # https://qiita.com/t_fuki/items/7cd50de54d3c5d063b4a#%E3%83%9D%E3%83%A9%E3%83%BC%E3%83%89%E3%83%AD%E3%83%BC%E7%B4%A0%E5%9B%A0%E6%95%B0%E5%88%86%E8%A7%A3%E6%B3%95%E3%81%AE%E3%82%A2%E3%83%AB%E3%82%B4%E3%83%AA%E3%82%BA%E3%83%A0 def gcd(a, b): while a: a, b = b%a, a return b def is_prime(n): if n == 2: return 1 if n == 1 or n%2 == 0: return 0 m = n - 1 lsb = m & -m s = lsb.bit_length()-1 d = m // lsb test_numbers = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37] for a in test_numbers: if a == n: continue x = pow(a,d,n) r = 0 if x == 1: continue while x != m: x = pow(x,2,n) r += 1 if x == 1 or r == s: return 0 return 1 def find_prime_factor(n): if n%2 == 0: return 2 m = int(n**0.125)+1 for c in range(1,n): f = lambda a: (pow(a,2,n)+c)%n y = 0 g = q = r = 1 k = 0 while g == 1: x = y while k < 3*r//4: y = f(y) k += 1 while k < r and g == 1: ys = y for _ in range(min(m, r-k)): y = f(y) q = q*abs(x-y)%n g = gcd(q,n) k += m k = r r *= 2 if g == n: g = 1 y = ys while g == 1: y = f(y) g = gcd(abs(x-y),n) if g == n: continue if is_prime(g): return g elif is_prime(n//g): return n//g else: return find_prime_factor(g) def factorize(n): res = {} while not is_prime(n) and n > 1: # nが合成数である間nの素因数の探索を繰り返す p = find_prime_factor(n) s = 0 while n%p == 0: # nが素因数pで割れる間割り続け、出力に追加 n //= p s += 1 res[p] = s if n > 1: # n>1であればnは素数なので出力に追加 res[n] = 1 return res # 高速約数列挙 def divisors(num): factors = factorize(num) divs = [1] for p in factors: e = factors[p] if e > 0: k = len(divs) #それまでの素因数積、つまり約数、の数 for i in range(e*k): divs.append(divs[-k]*p) #なぜans[-k]なのか、どんどんappendするので[-k]で前の約数にかけていく return divs # 高速約数カウント def divisor_count(num): factors = factorize(num) count = 1 for p in factors: e = factors[p] count *= (e+1) return count N = int(input()) divs = divisors(N) ans = -1 for d1 in divs: if d1**2 > N: break d2 = N//d1 if d1%2 == d2%2 and d2-d1 > 0: ans = 1 print(ans)