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 ATCODER

implicit none
integer(8) :: n,s,f,n_d,t,i,j,v1,v10,b
real(8) :: n_root

read(*,*) n,s

if(n<s)then
  write(*,'(i0)') -1
  stop
end if

if(n == s)then
  write(*,'(i0)') n+1
  stop
end if

n_root = real(n)**(0.5)
n_d = int(n_root)

do i = 2, n_d
  f = 0
  t = n
  do
    f = f + mod(t,i)
    if(t < i)exit
    t = t/i
  end do
  if(f == s)then
    write(*,'(i0)') i
    stop
  end if
end do

do v10 = 1, n_d
  if(mod(n-s,v10) /= 0)cycle
  b = (n-s)/v10 + 1
  f = mod(n,b) + n/b
  if(f == s)then
    write(*,'(i0)') b
    stop
  end if
end do

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

contains
!!!!!! サブルーチンは↓

!!!!!!

END PROGRAM ATCODER