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;

#define rep(i,a,b) for(int i=int(a);i<=int(b);++i)
#define nrep(i,a,b) for(int i=int(a);i>=int(b);--i)
#define trv(i, begin, end) for (__typeof(end) i = (begin) - ((begin) > (end)); i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
#define watch(x) cout<<"\n"<<#x<<" is "<<x<<"\n"
#define fast ios_base::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL);
#define ZERO(box) memset(box, 0, sizeof(box));
#define pb push_back
#define mt make_tuple
#define eb emplace_back
#define chkbit(s, b) (s & (1<<b))
#define leadzero(a) __builtin_clz(a) // count leading zeroes
#define trailzero(a) __builtin_ctz(a) // count trailing zeroes
#define bitcount(a) __builtin_popcount(a) // count set bits (add ll)
typedef long long ll;
typedef pair<int,int> pi;
typedef vector<int> vi;
typedef vector<vi> vvi;
const int mod=1e9+7;

bool check(ll n,ll b,ll s){
    ll sm=0;
    while(n){
        sm += n%b;
        n/=b;
    }
    return sm==s;
}

void solve(){
	ll n,s;
    cin>>n>>s;
    // base cases
    if(s>n){
        cout<<-1;
        return;
    }
    if(n==s){
        cout<<n+1;
        return;
    }
    ll b = 9e18;
    for(ll i=2;i*i<=n;i++){
        ll z = i;
        if(check(n,i,s)){
            cout<<i;
            return;
        }
    }
    // not any ans yet so let n=pb+q;
    ll p,q;
    for(p=1;p*p<=n;p++){
        q = s-p;
        if((n-q)%p ==0){
            b = min(b,(n-q)/p);
        }
    }
    if(b == 9e18){
        cout<<"-1";
    } else cout<<b;

}


int main(){
	int t=1;
	//cin>>t;
	while(t--){
		solve();	
	}	
}
/*input
99999999977
49999999989
*/