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 (1.15.1) (judge verdict: Accepted)

use std::io::*;
use std::str::FromStr;

fn read<T: FromStr>() -> T {
    let input = stdin();
    let input = input.lock();
    let token: String = input
            .bytes()
            .map(|c| c.expect("failure") as char)
            .skip_while(|c| c.is_whitespace())
            .take_while(|c| !c.is_whitespace())
            .collect();
    token.parse().ok().expect("error")
}

// 桁和関数の実装
fn digit_sum(n: i64, b: i64) -> i64 {
    if n < b {
        n
    } else {
        digit_sum((n as f64/b as f64).floor() as i64, b) + (n % b)
    }
}

fn main() {
    let n: i64 = read();
    let s: i64 = read();
    let mut ans: i64 = -1;

    if n == 1 {
        if s == 1 {
            ans = 2;
        }
    } else {

        let sqrtn = (n as f64).sqrt().ceil() as i64;

        // sqrt(n)+1から1までを先頭桁として全探索
        for p in (1..sqrtn+1).rev() {
            if (n-s+p) % p == 0 {
                let b: i64 = (n-s+p) / p;
                if b > 1 {
                    if digit_sum(n,b) == s {
                        ans = b;
                        break;
                    }
                }
            }
        }

        // 2からsqrt(n)まで全探索
        for b in 2..(sqrtn + 1) {
            if digit_sum(n, b) == s {
                ans = b;
                break;
            }
        }

        // n == s なら自分自身+1の時が常に最小
        if n == s {
            ans = n + 1;
        }
    }

    println!("{}",ans);
}