# 単調性はないので二分探索は無理
# 実験するとわかるのだがNの約数dに対して、d-1
# さらにd-1の約数、がpになりうる

def base_k(num, k):
    result = set()
    while num > 0:
        amari = num%k
        result.add(amari)
        num -= amari
        num //= k
    #print(result)
    return len(result)

def divisors(n):
    lower_divisors , upper_divisors = [], []
    i = 1
    while i*i <= n:
        if n % i == 0:
            lower_divisors.append(i)
            if i != n // i:
                upper_divisors.append(n//i)
        i += 1
    return lower_divisors + upper_divisors[::-1]

N = int(input())
if N <= 2:
    print(N+1)
    exit()

p_candidates = set()
N_divs = divisors(N)
for d in N_divs:
    d1_divs = set(divisors(d-1))
    p_candidates |= d1_divs
p_candidates.discard(0)
p_candidates.discard(1)
p_candidates = sorted(list(p_candidates))
#print(p_candidates)
for p in p_candidates:
    calc = base_k(N, p)
    #print('p', p, 'calc', calc)
    if calc==1:
        print(p)
        break