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>
using namespace std;


bool check(long long n, long long s, long long b){
    if(b<2) return false;
    long long sum = 0;
    while(n>0){
        sum+=n%b;
        n/=b;
    }
    return sum==s;
}


int main(){
    long long n, s, maxsum=-1;
    cin>>n>>s; if(n==s){ cout<<n+1; return 0;}
    /*for(int i =2; i< n; i++){
        long long buf=n;
        long long sum=0;
        while(buf>0){
            sum+=buf%i;
            cout<<buf%i<<" ";
            buf/=i;
        }
        //maxsum = max(sum, maxsum);
        cout<<" separator "<<i<<" "<<sum;
        if(sum==s) cout<<" equals ";
        cout<<endl;
    }*/
    long long answer = 10000000000000;
    for(long long i=1; i<(long long)sqrt(n)+10000; i++){
        if(check(n, s, i)) answer = min(answer, i);
        long long a=i, b=s-i; if(b<0) continue;
        long long base = (n-b)/a;
        if(check(n, s, base)) answer = min(answer,base);
    }
    cout<<(answer!=10000000000000?answer: -1);
    return 0;
}