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>
#include<vector>
#include<cstring>
#include<queue>
#include<cmath>
#include<algorithm>
using namespace std;
long long n,s;
long long f(long long x,long long y)
{
	long long re=0;
	while(x!=0)
	{
		re+=x%y;
		x/=y;
	}
	return re;
}
int main()
{
	scanf("%lld %lld",&n,&s);
	
	if(s==1)
	{
		printf("%lld\n",n);
		return 0;
	}
	long long e=n-s;
	if(e<0)
	{
		printf("-1\n");
		return 0;
	}
	for(long long i=1;i*i<=e;i++)
	{
		if(e%i==0)
		{
			long long ans=i+1,k=e/i;
			if(f(n,ans)==s)
			{
				printf("%lld\n",ans);
				return 0;
			}
		}
	}
	printf("-1\n");
}