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: Wrong Answer)

#include <stdio.h>
#include <math.h>
#include <limits.h>

typedef long long ll;

ll min(ll a,ll b){
	if(a<b){return a;}
	else{return b;}
}

ll f(ll b,ll n){

	ll ret;

	if(n<b){ret=n;}
	else{ret=f(b,n/b)+(n%b);}

	return ret;

}

int main(){

	ll n,s;

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

	ll ans=LLONG_MAX;

	ll b;

	if(n-s>=0ll){
		for(b=2ll;b<=(ll)sqrt((double)n-(double)s)+2ll;b+=1ll){
			if((n-s)%(b-1ll)==0ll){
				if(f(b,n)==s){
					ans=min(ans,b);
				}
				if(((n-s)/(b-1ll))+1ll>=2ll){
					if(f(((n-s)/(b-1ll))+1ll,n)==s){
						ans=min(ans,((n-s)/(b-1ll))+1ll);
					}
				}
			}
		}
	}

	if(ans<LLONG_MAX){printf("%lld\n",ans);}
	else{printf("-1\n");}

	return 0;
}