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)

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

flag = False
if s > n:
  print(-1)
  exit()
if s > n/2:
  print(-1)
  exit()
  
if s == n:
  print(n+1)
  exit()

def chk(b, n):
  if n//b == 0:
    return n
  else:
    return chk(b,n//b) + n%b

for b in range(2, int(n**0.5)+1):
  if chk(b, n) == s:
    print(b)
    flag = True
    exit()

for i in range(int(n**0.5), 0, -1):
  b = (n-s)//i + 1
  if chk(b, n) == s:
    print(b)
    flag = True
    exit()

if flag == False:
  print(-1)