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: Wrong Answer)

import math
import sys
sys.setrecursionlimit(10**5)
n = int(input())
s = int(input())

if n < s:
    print (-1)
    exit()

def f(b, n):
    if n < b:
        return n
    else:
        return f(b, n//b) + n%b

if s == n:
    print (n+1)
    exit()

if n == 1:
    print (-1)
    exit()

for i in range(2, int(math.sqrt(n))+2):
    if f(i, n) == s:
        print (i)
        exit()

for p in range(1, int(math.sqrt(n))+2):
    b = (n-s)//p + 1
    # print (b)
    if b != 1 and f(b, n) == s:
        print (b)
        exit()
print (-1)

# for i in range(2, 11):
#     print (i, f(i,10))