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;
typedef long long ll;
typedef pair<int,int> pii;
#define rep(i,n) for(ll i=0;i<(ll)(n);i++)
#define all(a)  (a).begin(),(a).end()
#define vi vector<int>
#define pb push_back
//#define INF 999999999
#define INF (1LL<<59)

#define int ll

int f(int b,int n){
	if(n<b)return n;
	else{
		return (f(b,floor(n/b))+(n%b));
	}
}

set<ll> prime_factorization(ll n){
	set<ll> ret;
	
	for(ll div=2;div<=1+sqrt(n);div++){
		while(n%div==0){
			ret.insert(div);
			n/=div;
		}
	}
	if(n!=1)ret.insert(n);
	return ret;
}



signed main(){
	int n,s;
	cin>>n>>s;
	if(n==s){
		cout<<n+1<<endl;
		return 0;
	}
	
	for(int i=2;i*i<=n;i++){
		if(f(i,n)==s){
			cout<<i<<endl;
			return 0;
		}
	}

	
	
	set<ll> st = prime_factorization(n-s);
	st.insert(n-s);
	int ans = INF;

	for(auto &elm:st){
		int B = elm+1;
		int p = (n-s)/(B-1);
		int q = s-p;
		if(q<0)continue;
		
		if(p*B+q==n)ans = min(ans,B);
	}
	
	if(ans!=INF)cout<<ans<<endl;
	else cout<<-1<<endl;
}