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 <bits/stdc++.h>
#define int long long
using namespace std;
int n,s;
bool f(int b){
    int ans = 0;
    int cur = n;
    while (cur){
        ans += cur%b;
        cur /= b;
    }
    return ans == s;

}
main(){
    scanf("%lld%lld",&n,&s);
    if (s == n){
        printf("%lld",n+1); return 0;
    }
    if (s > n){
        printf("-1\n");
        return 0;
    }
    for (int i = 2; i*i <= n; i++){
        if (f(i)){
            printf("%lld",i);
            return 0;
        }
    }
    for (int i = 2; i*i <= n; i++){
        if (f(i)){
            printf("%lld",i);
            return 0;
        }
    }
    int ans = -1;
    for (int i = 1; i*i <= n; i++){
        if ((n-s)%i == 0){
            int k = (n-s)/i + 1;
            if (f(k)){
                if (ans == -1 || ans > k) ans = k;
            }
        }
    }
    printf("%lld",ans);
}