Problem:
Score : 500 points

Problem Statement

For integers b (b \geq 2) and n (n \geq 1), let the function f(b,n) be defined as follows:

f(b,n) = n, when n < b

f(b,n) = f(b,\,{\rm floor}(n / b)) + (n \ {\rm mod} \ b), when n \geq b

Here, {\rm floor}(n / b) denotes the largest integer not exceeding n / b,
and n \ {\rm mod} \ b denotes the remainder of n divided by b.

Less formally, f(b,n) is equal to the sum of the digits of n written in base b.
For example, the following hold:

f(10,\,87654)=8+7+6+5+4=30

f(100,\,87654)=8+76+54=138

You are given integers n and s.
Determine if there exists an integer b (b \geq 2) such that f(b,n)=s.
If the answer is positive, also find the smallest such b.

Constraints

1 \leq n \leq 10^{11}

1 \leq s \leq 10^{11}

n,\,s are integers.

Input

The input is given from Standard Input in the following format:

n
s

Output

If there exists an integer b (b \geq 2) such that f(b,n)=s, print the smallest such b.
If such b does not exist, print -1 instead.

Sample Input 1

87654
30

Sample Output 1

10

Sample Input 2

87654
138

Sample Output 2

100

Sample Input 3

87654
45678

Sample Output 3

-1

Sample Input 4

31415926535
1

Sample Output 4

31415926535

Sample Input 5

1
31415926535

Sample Output 5

-1

Solution in Python (3.4.3) (judge verdict: Accepted)

import math

N = int(input())
S = int(input())

def f(B, N):
    X, res = N, 0
    while 0 < X:
        res += X % B
        X //= B
    return res


if N == S:
    print(N + 1)
    exit()

i = 2
while i * i <= N:
    if f(i, N) == S:
        print(i)
        exit()
    i += 1

sqrtN = math.floor(math.sqrt(N))
i = 1
ans = -1
while i * i < N:
    if (N - S) % i != 0:
        i += 1
        continue
    b = (N - S) // i + 1
    if sqrtN < b and f(b, N) == S:
        ans = b
    i += 1

print(-1 if ans == -1 else ans)