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: Memory Limit Exceeded)

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