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 <cstdio>
#include <cmath>
using namespace std;

int n, s;

inline bool is_jz(int x, int b) {
	int sum = 0;
	while (x)
		sum += x % b, x /= b;
	return sum == s;
}

int main() {
	scanf("%d %d", &n, &s);
	if(n == s) {
		printf("1");
		return 0;
	}
	int sqr = sqrt(n);
	for (int i = 2; i <= sqr; ++i)
		if (is_jz(n, i)) {
			printf("%d", i);
			return 0;
		}
	for (int i = sqr; (n - s) / i + 1 > i; --i)
		if ((n - s) % i == 0) {
			printf("%d", (n - s) / i + 1);
			return 0;
		}
	printf("-1");
	return 0;
}