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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<cstdio>
#define LL long long 
LL n,s;
LL f(LL x,LL b)
{
	LL res=0;
	while(x)
	{
		res+=x%b;
		x/=b;
	}
	return res;
}
int main()
{
	scanf("%lld%lld",&n,&s);
	if(s>n)
	{
		printf("-1\n");
		return 0;
	}
	if(s==n)
	{
		printf("%lld\n",n+1);
		return 0;
	}
	for(LL i=2ll;i*i<=n;i++)
		if(f(n,i)==s)
		{
			printf("%lld\n",i);
			return 0;
		}
	for(LL i=1ll;i*i<=n-s;i++)
		if((n-s)%i==0)
		{
			LL b=(n-s)/i+1;
			if(n%b+n/b==s)
			{
				printf("%lld\n",b);
				return 0;
			}
		}
	printf("-1\n");
}