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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <math.h>

long long getsum(long long b, long long n){
    if (n<b) return n;
    else return getsum(b, n/b) + n % b;
}

int p_max(long long n, long long s){
    int sq = sqrt(n);
    if (s < sq) return s;
    else if (sq*sq < n) return sq;
    else return sq-1;
}

int main(){
    
    long long n, s;
    
    scanf("%lld %lld", &n, &s);
    
    if (s == n){
        printf("%lld\n", n+1);
        return 0;
    }
    if (s > n){
        printf("%d\n", -1);
        return 0;
    }
    
    for (int b=2; b<=sqrt(n); b++){
        if (getsum(b, n) == s){
            printf("%d\n", b);
            return 0;
        }
    }
    
    for (int p=p_max(n, s); p>=1; p--){
        if ((n-s) % p == 0 && s-p < (n-s)/p+1){
            printf("%lld\n", (n-s)/p+1);
            return 0;
        }
    }
    
    printf("%d\n", -1);
    return 0;
}