# 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_n(num_10,n):
    str_n = ''
    while num_10:
        if num_10%n>=10:
            return -1
        str_n += str(num_10%n)
        num_10 //= n
    return int(str_n[::-1])

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())
p_candidates = set()
N_divs = divisors(N)
for d in N_divs:
    p_candidates.add(d-1)
    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_n(N, p)
    if calc == -1:
        continue
    S = set()
    for c in str(calc):
        S.add(c)
    #print('p', p, 'calc', calc, 'S', S)
    if len(S)==1:
        print(p)
        break