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 fi first
#define se second
using namespace std;
typedef long long ll;
const double PI=acos(-1.0);
const double eps=1e-6;
const ll inf=1e18;
const int mod=1e9+7;
const int maxn=5e5+10;
ll n,s;
ll f(ll x,ll b){
	ll ret=0;
	while(x){
		ret+=x%b;
		x/=b;
	}
	return ret;
}
int main(){
    ios::sync_with_stdio(false);
    //cout<<(1ll<<26)<<endl;
   	cin>>n>>s;
   	for(ll i=2;i*i<=n;i++){
   		if(f(n,i)==s){
   			cout<<i<<endl;
   			return 0;
   		}
   	}
   	ll k=n-s;
   	ll ans=inf;
   	for(ll i=1;i*i<=k;i++){
   		if(k%i==0){
   			if(f(n,k/i+1)==s) ans=min(ans,k/i+1);
   			if(f(n,i+1)==s) ans=min(ans,i+1);
   		}
   	}
   	if(ans==inf) ans=-1;
   	cout<<ans<<endl;
   	/*p*b+q==n
   	p+q==s
   	(n-s)/p==(b-1);*/
   /*	q=s-p;
	k/(k/i);*/
    return 0;
}