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 <bits/stdc++.h>
#define pb push_back
#define tam 310000
#define offset 100000
#define se second
#define fi first
#define forr(i,p,n) for(ll i=p;i<n;i++)
using namespace std;
typedef long long ll;
const ll MOD=1e9+7;
ll mult(ll a,ll b, ll p=MOD){return ((a%p)*(b%p))%p;}
ll add(ll a, ll b, ll p=MOD){return (a%p + b%p)%p;}
typedef pair<ll,ll> ii;
ll f(ll num,ll base)
{
    if (num==0)
        return 0;
    //cout<<num%base<<" ";
    return num%base+f(num/base,base);
}
int main()
{
    ll n,s;
    cin>>n>>s;
    if (s==n)
    {
        cout<<n+1<<endl;
        return 0;
    }
    forr(i,2,1e6)
    if (f(n,i)==s)
        {
            cout<<i<<endl;
            return 0;
        }
    ll val=1e6,minn=1e18;
    while(val<=n)
    {
        ll iz=val,der=n+1,mid;
        while(iz+1<der)
        {
            mid=(iz+der)/2;
            if (n/mid==n/val)
                iz=mid;
            else
                der=mid;
        }
        ll fo=f(n,val);
        ll X=n/val;
        //cout<<val<<" "<<iz<<" "<<endl;
        if (fo>=s&& (fo-s)%X==0&& iz-val<=(fo-s)/X)
            minn=min(minn,val+(fo-s)/X);
        val=der;

    }
    if (minn==1e18)
    cout<<-1<<endl;
    else
    cout<<minn<<endl;
}