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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

import math 
import bisect
import sys
def f(b,n):
    m=math.floor(n/b)
    if n>b:
        return f(b,m) + n%b
    else:
        return n
n=int(input())
s=int(input())
for i in range(2,int(math.sqrt(n))+1):
    x=f(i,n)
    if x==s:
        print(i)
        sys.exit()
if n-int(math.sqrt(n))<2:
    l=2
else:
    l=n-int(math.sqrt(n))
for i in range(n+1,l-1,-1):
    x=f(i,n)
    if x==s:
        print(i)
        sys.exit()
print(-1)