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 Java (OpenJDK 11.0.6) (judge verdict: Runtime Error)

import java.io.*;
import java.util.*;
import static java.lang.System.out;

public class Main {
    static MyReader in = new MyReader();

    public static void main(String[] args){
        long n = in.l();
        long s = in.l();
        out.println(solve(n, s));
    }

    static long solve(long n, long s) {
        if (s == n) {
            return n + 1;
        } else {
            double sqrtN = Math.sqrt(n);
            for (int b = 2; b <= sqrtN; b++) {
                if (s == f(b, n)) {
                    return b;
                }
            }
            for (long p = 1; p < sqrtN; p++) {
                long b = (n - s) / p + 1;
                if (s == f(b, n)) {
                    return b;
                }
            }
            return -1;
        }
    }

    static long f(long b, long n) {
        long sum = 0;
        while (n > 0) {
            sum += n % b;
            n /= b;
        }
        return sum;
    }
}

class MyReader extends BufferedReader {
    char[] cbuf = new char[1024];
    int head = 0;
    int tail = 0;

    MyReader() {
        super(new InputStreamReader(System.in));
    }

    char next() {
        if (head == tail) {
            try {
                tail = read(cbuf, 0, cbuf.length);
            } catch (IOException e) {}
            head = 0;
        }
        return cbuf[head++];
    }

    void back() {
        head--;
    }

    boolean minus() {
        boolean minus;
        while (true) {
            char c = next();
            if (c != ' ' && c != '\n' && c != '\r') {
                if (!(minus = c == '-')) back();
                return minus;
            }
        }
    }

    void skip() {
        while (true) {
            char c = next();
            if (c != ' ' && c != '\n' && c != '\r') {
                back();
                break;
            }
        }
    }

    char[] s(final int N) {
        skip();
        char[] s = new char[N];
        for (int i = 0; i < s.length; i++) {
            s[i] = next();
        }
        return s;
    }

    int i() {
        boolean minus = minus();
        int n = 0;
        while (true) {
            int k = next() - '0';
            if (k < 0 || 9 < k) break;
            n = 10 * n + k;
        }
        return minus ? -n : n;
    }

    int[] ii(final int N) {
        int[] a = new int[N];
        for (int j = 0; j < a.length; j++) a[j] = i();
        return a;
    }

    long l() {
        boolean minus = minus();
        long n = 0;
        while (true) {
            int k = next() - '0';
            if (k < 0 || 9 < k) break;
            n = 10 * n + k;
        }
        return minus ? -n : n;
    }
}