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: Runtime Error)

#include <stdio.h>

#define lln long long int

lln digit_sum(lln b, lln n){
	if(n < b){
		return n;
	}
	else{
		return digit_sum(b, n / b) + n % b;
	}
}

int main(){
	lln n, s, b, p;
	scanf("%lld%lld", &n, &s);
	if(n < s){
		printf("-1\n");
		return 0;
	}
	if(n == s){
		printf("%lld\n", n + 1);
		return 0;
	}
	for(b = 2; b * b <= n; b++){
		if(digit_sum(b, n) == s){
			printf("%lld\n", b);
			return 0;
		}
	}
	for(p = b - 1; p >= 1; p--){
		b = (n - s) / p + 1;
		if(digit_sum(b, n) == s){
			printf("%lld\n", b);
			return 0;
		}
	}
	printf("-1\n");
	return 0;
}