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>
using namespace std;
using ll=long long;
using vb=vector<bool>;
using vvb=vector<vb>;
using vd=vector<double>;
using vvd=vector<vd>;
using vi=vector<int>;
using vvi=vector<vi>;
using vl=vector<ll>;
using vvl=vector<vl>;
using pll=pair<ll,ll>;
using tll=tuple<ll,ll>;
using vs=vector<string>;
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define rep(i,n) range(i,0,n)
#define rrep(i,n) for(int i=(n)-1;i>=0;i--)
#define range(i,a,n) for(int i=(a);i<(n);i++)

#define LINF    	((ll)1ll<<60)
#define INF     	((int)1<<30)
#define EPS     	(1e-9)
#define MOD     	(1000000007ll)
#define fcout(a)	cout<<setprecision(a)<<fixed
#define fs 			first
#define sc			second
#define PI			(3.1415926535897932384)

int dx[]={1,0,-1,0,1,-1,-1,1},dy[]={0,1,0,-1,1,1,-1,-1};
template<class S,class T>ostream&operator<<(ostream&os,pair<S,T>p){os<<"["<<p.first<<", "<<p.second<<"]";return os;};
template<class S>auto&operator<<(ostream&os,vector<S>t){bool a=1; for(auto s:t){os<<(a?"":" ")<<s; a=0;} return os;}
template<class T> void chmax(T &a, const T &b){if(a<b)a=b;}
template<class T> void chmin(T &a, const T &b){if(a>b)a=b;}
void YN(bool b){cout<<(b?"YES":"NO")<<endl;}
void Yn(bool b){cout<<(b?"Yes":"No")<<endl;}
void yn(bool b){cout<<(b?"yes":"no")<<endl;}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	ll n,nsub,s,sub;
	cin>>n>>s;
	ll i;
	for(i=2;i*i<=n;i++){
		nsub=n;
		sub=0;
		while(i<=nsub){
			sub+=(nsub%i);
			nsub/=i;
		}
		sub+=nsub;
		if(sub==s){
			cout<<i<<endl;
			return 0;
		}
	}
	for(;i>0;i--){
		if((n-s)>=i&&(n-s)%i==0){
			ll b=(n-s)/i+1;
			if(s>=i&&s-i==n-i*b){
				cout<<b<<endl;
				return 0;
			}
		}
	}
	if(n==s){
		cout<<n+1<<endl;
		return 0;
	}
	cout<<-1<<endl;
}