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)

//#pragma GCC optimize(\"Ofast,no-stack-protector\")
//#pragma GCC target(\"sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx\")
//#pragma GCC target(\"avx,tune=native\")
// Neil

#include<bits/stdc++.h>

using namespace std;

typedef complex<double> base;
typedef long double ld;
typedef long long ll;

#define endl "\n"
#define pb push_back
#define pii pair<int,int>
#define pll pair< ll , ll >
#define vi vector<int>
#define vvi vector< vi >
const int MAXN=(int)(2e5+5);
const ll MOD=(ll)(1e9+7);
int t;

ll func(ll b, ll n){
    if(n<b) return n;
    else return func(b, n/b) + n%b;
}
int main()
{
    ios_base::sync_with_stdio(0);cin.tie(0);
    ll n,s;
    cin>>n>>s;
    if(n==s){
        cout<<n+1<<endl;
        return 0;
    }
    ll ans = -1;
    for(int b = 2;b*b<=n;b++){
        if(func(b, n) == s){
            cout<<b<<endl;
            return 0;
        }
    }

    // n = pb + q
    // s = p + q
    for(ll p = 1;p*p<n;p++){
        ll b = (n-s)/s + 1;
        if(b>=2 and func(b, n) == s){ ans = b; break;} 
    }
    cout<<ans<<endl;
    return 0;
}