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 (3.8.2) (judge verdict: Wrong Answer)

import math

n = int(input())
s = int(input())


r = 0

if n == s:
    print(n + 1)
    exit()
elif n < s:
    print(-1)
    exit()

sq = int(math.sqrt(n))


# 2 <= d <= √n
for i in range(2, sq+1):
    nt = n
    st = 0

    while nt > 0:
        st += nt % i
        nt //= i

    if st == s:
        print(i)
        exit()
    
# √n < d <= n
for i in range(sq-1, 0, -1):
    b = (n - s) // i + 1
    
    st = n % b + n // b

    if st == s:
        print(b)
        exit()

print(-1)