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 Fortran (gfortran v4.8.4) (judge verdict: Wrong Answer)

program main
  implicit none
  integer(8) :: n, s, sqrt_n, b, p
  read(*, *) n
  read(*, *) s

  if (n < s) then
    write(*, "(i0)") -1
    stop 0
  else if (n == s) then
    write(*, "(i0)") n + 1
    stop 0
  end if

  sqrt_n = integer_sqrt(n)

  do b = 2, sqrt_n
    if (f(b, n) == s) then
      write(*, "(i0)") b
      stop 0
    end if
  end do

  do p = 1, sqrt_n
    b = (n - s) / p + 1
    if (b >= 2) then
      if (f(b, n) == s) then
        write(*, "(i0)") b
        stop 0
      end if
    end if
  end do

  write(*, "(i0)") -1

  contains

  function integer_sqrt(n)
    implicit none
    integer(8), intent(in) :: n
    integer(8) :: integer_sqrt, i
    if (n < 0) stop 101
    do i = 0, n + 1
      if (i ** 2 > n) then
        integer_sqrt = i - 1
        exit
      end if
    end do
  end function integer_sqrt

  recursive function f(b, n) result(res)
    implicit none
    integer(8), intent(in) :: b, n
    integer(8) :: res
    if (n < b) then
      res = n
    else
      res = f(b, n / b) + mod(n, b)
    end if
  end function f
end program main