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<iostream>
#include<cstdio>
#include<map>
#include<cmath>
#include<sstream>
#include<cstring>
#include<algorithm>
#include<string>
#define LL long long
using namespace std;
LL n,s,b;
bool check(LL n, LL b)
{
    long long res = 0;
    while(n > 0)
    {
        res += n%b;
        n/=b;
    }
    return res == s;
}
int main()
{
    //int n,s,b;
    cin >> n >> s;
    LL temp = (LL)sqrt(n)+1;
    for(LL i = 2; i <= temp; i ++)
    {
        if(check(n,i))
        {
            cout << i << endl;
            return 0;
        }
    }
    if(s == n)
    {
        cout << n+1 << endl;
        return 0;
    }
    if(s > n)
    {
        cout << "-1" << endl;
        return 0;
    }
    LL temp1 = n - s, temp2 = temp1/temp + 1;
    for(int i = temp2; i >= 1; i --)
    {
        if(!(temp1%i) && s - i <= temp1/i && s >= i && i<= temp1/i)
        {
            cout << temp1/i + 1 << endl;
            return 0;
        }
    }
    cout << "-1" << endl;
    return 0;
}