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: Accepted)

#include <stdio.h>
#include <math.h>
typedef long long LL;

int main() 
{
	LL p, q, k, i, t, s, x, y;
	
	scanf("%lld%lld", &p, &q);
	if(q > p){
		puts("-1");
		return 0;
	}
	if(q == p){
		printf("%lld\n", p + 1);
		return 0;
	}
	for(i = 2, k = sqrt(p); i <= k; i++){
		t = p, s = 0;
		while(t){
			s += t % i;
			t /= i;
		}
		if(s == q){
			printf("%lld\n", i);
			return 0;
		}
	}
	for(t = p - q, i = sqrt(t); i > 0; i--){
		if(t % i) continue;
		x = t / i + 1;
		y = p - x * i;
		if(y >= 0 && y < x){
			printf("%lld\n", x);
			return 0;
		}
	}
	puts("-1");
	
	return 0;
}