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 <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <queue>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

typedef long long ll;
const int INF = 1e9;
using namespace std;


ll n,s;

ll f(ll b,ll n){
    if(n<b) return n;
    return f(b,floor(n/b))+n%b;
}
void solve(){
    if(s>n){
        printf("-1\n");
        return;
    }
    else if(s==n){
        printf("%lld\n",n+1);
        return;
    }
    else{
        ll arieru_m = f(n/2+1,n);
        if(arieru_m<s){
//        cout << arieru_m <<"よりでかいのでむり"<<endl;
            printf("-1\n");
            return;
        }
        if(arieru_m==s){
            printf("%lld\n",n/2+1);
            return;
        }
//        ひたすらさがす
        if(s==1&&n%2){
            printf("%lld\n",n);
            return;
        }
        for(ll b = 2; b<=n/2;b++){
            if(s==f(b,n)){
                printf("%lld\n",b);
                return;
            }
        }
    }
    printf("-1\n");
}

int main() {
    scanf("%lld",&n);
    scanf("%lld",&s);
    solve();
    return 0;
}