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 C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
int main(){
  long long n,m,i,j,s;
  scanf("%lld %lld",&n,&m);
  if(n==m){
    printf("%lld\n",n+1);
    return 0;
  }
  for(i=2;i*i<=n;i++){
    s=0;
    for(j=n;j;j/=i)s+=j%i;
    if(s==m){
      printf("%lld\n",i);
      return 0;
    }
  }
  if((i-1)*(i-1)==n)i--;
  for(i--;i;i--){
    j=n/(i+1)+1;
    s=n/j+n%j;
    if(m>s)continue;
    if((s-m)%i==0&&s-m<=n%j&&j+(s-m)/i-1){
      printf("%lld\n",j+(s-m)/i);
      return 0;
    }
  }
  printf("-1\n");
  return 0;
}