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 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>

long long N, S;
long long ans;

long long f(long long b, long long n)
{
	return (b > n) ? n : f(b, n / b) + n % b;
}

long long func(void)
{
	long long t;
	long long i, j;
	long long k = 320000;

	if (S > N / 2 + (N & 1LL)) return -1LL;
	if (0 == N % S) return N / S;

	if (k > N) k = N;
	for (i = 2LL; i <= k; ++i)
	{
		if (f(i, N) == S) return i;
	}

	for (i = N / k; i > 1; --i)
	{
		t = (S - N % i) / i;
		if (0 == t && t <= (N / i - N / (i + 1)))
		{
			return t + N / i;
		}
	}

	return N + 1 - S;
}

int main(void)
{
	scanf("%lld %lld", &N, &S); //printf("%lld\n", f(N / 11, N));
	printf("%lld\n", func());

	return 0;
}