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: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <list>
#include <cmath>
#include <stack>
#include <iomanip>

using namespace std;
typedef long long ll;
using Pll = pair<ll,ll>;

const ll INF = 1LL<<60;
const ll MOD = 1000000007;

//cout << fixed << setprecision(10);


ll digit_sum(ll b,ll N){
    if(N<b) return N;
    else return digit_sum(b,floor((double)N/b)) + (N%b);
}

int main() {
    ll N,S;
    cin >> N >> S;
    if(N<S){
        cout << -1 << endl;
        return 0;
    }
    for(int i=2;i<=floor(sqrt(N));i++){
        if(digit_sum(i,N)==S){
            cout << i << endl;
            return 0;
        }
    }

    for(int i=floor(sqrt(N));i>0;i--){
        if((N-S)%i!=0 || (N-S)/i==0) continue;
        else{
            if(digit_sum((N-S)/i+1,N)==S){
                cout << (N-S)/i+1 << endl;
                return 0;
            }
        }
    }

    cout << -1 << endl;
    return 0;
}