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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#define REP(i, n) for(int i = 0; i < n; i++)
#define REPR(i, n) for(int i = n; i >= 0; i--)
#define FOR(i, m, n) for(int i = m; i < n; i++)
#define ALL(obj) (obj).begin(), (obj).end()
#define INF 1e9
#define LINF 1e18
typedef long long ll;

int digit(ll b, ll n){
    ll num = n;
    int ans = 0;
    while(num>0){
        ans += num%b;
        num /= b;
    }
    return ans;
}

int main() {
    ll n,s; cin >> n >> s;

    if(s==n){
        cout << n+1 << endl;
        return 0;
    }

    if(n<s){
        cout << -1 << endl;
        return 0;
    }

    FOR(b,2,sqrt(n)+1){
        if(digit(b,n)==s){
            cout << b << endl;
            return 0;
        }
    }

    for(ll p = sqrt(n)+1; p>=1; p--){
        if((n-s)%p == 0){
            ll b = (n-s)/p + 1;

            if(digit(b,n)==s){
                cout << b << endl;
                return 0;
            }

        }
    }

    cout << -1 << endl;
    return 0;
}