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: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

int main( int argc, char *argv[] )
{
  long long n,b;
  long long nn;
  long long sn;
  long long s;
  long long a;
  int i;

  scanf( "%lld%lld", &n, &s );

//  printf( "%lld %lld\n", n, s );

  if( n == s ){
    printf( "%lld\n", n+1 );
    return 0;
  }
  if( n < s ){
    printf( "%d\n", -1 );
    return 0;
  }
  for( b = 2; b <= n; b ++ ){
    nn = n;
    a = 0;
    while( nn ){
      a += nn%b;
      nn = nn/b;
    }
    if( a == s ){
      printf( "%lld\n",b );
      return 0;
    }
  }

  printf( "-1\n" );
  return 0;
}