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 Rust (1.42.0) (judge verdict: Accepted)

fn main() {
	let mut s = String::new();
	use std::io::Read;
	std::io::stdin().read_to_string(&mut s).unwrap();
	let mut s = s.split_whitespace();
	let n: i64 = s.next().unwrap().parse().unwrap();
	let s: i64 = s.next().unwrap().parse().unwrap();

	let ans = calc(n, s);
	println!("{}", ans);
}

fn calc(n: i64, s: i64) -> i64 {
	use num::integer::Roots;
	let sqrtn = n.sqrt();
	for b in 2..=sqrtn {
		let ds = f(b, n);
		if ds == s {
			return b;
		}
	}

	if n > s {
		for d in divisors(n - s) {
			let b = d + 1;
			let p = (n - s) / d;
			if b > sqrtn && p < b {
				let ds = p + n % b;
				if ds == s {
					return b;
				}
			}
		}
	}
	if n == s {
		return n + 1;
	}

	-1
}

fn f(b: i64, mut n: i64) -> i64 {
	let mut r = 0;
	while n > 0 {
		r += n % b;
		n /= b;
	}
	r
}

pub fn divisors<T>(n: T) -> impl std::iter::Iterator<Item = T>
where
	T: num::PrimInt,
{
	let n = if n < T::zero() { T::zero() - n } else { n };
	let mut d = T::one();
	let mut divs = vec![];
	std::iter::from_fn(move || {
		while d * d <= n {
			if n % d == T::zero() {
				let r = d;
				if d != n / d {
					divs.push(n / d);
				}
				d = d + T::one();
				return Some(r);
			}
			d = d + T::one();
		}

		divs.pop()
	})
}