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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <cstdio>
#include <functional>
#include <cmath>

using namespace std;

int main()
{
    long long N, S;
    scanf("%lld\n", &N);
    scanf("%lld\n", &S);

    if (N == S){
        printf("%lld", N + 1);
        return 0;
    }else if (N < S || S > (N + 1) / 2){
        puts("-1");
        return 0;
    }

    function<long long(long long, long long)> f = [&](long long b, long long n){
        if (n < b){
            return n;
        }else{
            return f(b, n / b) + (n % b);
        }
    };
    const long long sqrt_N = ceil(sqrt(N));
    for (long long b = 2; b <= sqrt_N; ++b){
        if (f(b, N) == S){
            printf("%lld\n", b);
            return 0;
        }
    }
    const long long sub = N - S;
    for (long long p = sqrt_N; p > 0; --p){
        if (sub % p) continue;
        const long long b = (sub / p) + 1;
        if (f(b, N) == S){
            printf("%lld\n", b);
            return 0;
        }
    }
    puts("-1");
}