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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

long long n, s, b;
inline long long f(long long b, long long n){
    if(n < b) return n;
    return f(b, n / b) + n % b;
}
int main()
{
    scanf("%lld%lld", &n, &s);
    if(s == 1) {printf("%lld", n); return 0;}
    if(s == n) {printf("%lld", n + 1); return 0;}
    for(long long b = 2; b * b <= n ; ++b){
        long long sum = f(b, n);
        if(sum == s){
            printf("%lld", b);
            return 0;
        }
    }
    long long m = n - s;
    for(long long d = 2; d * d <= n ; ++d){
        if(m % d == 0){
            long long b = m / d;
            long long sum = f(b, n);
            if(sum == s){
                printf("%lld", b);
                return 0;
            }
            b = d;
            sum = f(b, n);
            if(sum == s){
                printf("%lld", b);
            }
        }
    }
    printf("-1");
    return 0;
}