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 <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll get(ll x, ll y)
    {
    ll s = 0;
    while(x)
        {
        s += x%y;
        x/=y;
    }
    return s;
}
int main()
    {
    ll n,s;
    scanf("%lld%lld",&n,&s);
    if(n==1)
        {
        if(s==1)return printf("2\n")*0;
        return printf("-1")*0;
    }
    set<ll> pos;
    ll x;
    pos.insert(n+1);
    if(n!=1)
    pos.insert(n);
    if(n > s)
        {
        ll dif = n-s;
        
        for(x=2;x*x<=dif;x++)if(dif%x==0)
            {
            pos.insert(x);
            pos.insert(x+1);
            if(x*x!=dif){
                pos.insert(dif/x);
                pos.insert(dif/x+1);
            }
        }
        if(dif!=1){
        pos.insert(dif);
            pos.insert(dif+1);
        }
    }
    for(x=2;x*x<=n;x++)pos.insert(x);
    set<ll>::iterator I;
    for(I=pos.begin();I!=pos.end();I++){
     //   printf("checking for %lld\n",*I);
        if(get(n,*I)==s)return printf("%lld\n",*I)*0;
   
}
        return printf("-1\n")*0;
  //  if(s==1)return printf("%lld\n",n)*0;
    
}