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,j,n) for(int i=(int)(j);i<(int)(n);i++)
#define REP(i,j,n) for(int i=(int)(j);i<=(int)(n);i++)
#define per(i,j,n) for(int i=(int)(j);(int)(n)<=i;i--)
#define ll long long
#define ALL(a) (a).begin(),(a).end()
#define disup(A,key) distance(A.begin(),upper_bound(ALL(A),(int)(key)))
#define dislow(A,key) distance(A.begin(),lower_bound(ALL(A),(int)(key)))
#define pb push_back
#define mp make_pair
#define endl "\n"
using std::cin;
using std::cout;
using std::vector;
using std::pair;
using std::string;
using std::upper_bound;
using std::lower_bound;
using vi=vector<int>;
using vii=vector<vi>;
using pii=pair<int,int>;
const ll MOD=1e9+7;
const int MAX=1e5+1;
const ll INF=(1ll<<62)-1;
ll disit_sum(int b,int n){
    if(n<b) return n;
    else return disit_sum(b,n/b)+n%b;
}
signed main(){
    ll n,s; cin>>n>>s;
    ll ans=INF;
    REP(i,2,sqrt(n)){
        ll X=disit_sum(i,n);
        if(X==s) ans=std::min(ans,(ll)i);
    }
    REP(i,1,sqrt(n)){
        ll mod=s-i;
        if(mod<0) break;
        if(mod>n) continue;
        ll quo=n-mod;
        if(quo%i==0&&quo/i>mod) ans=std::min(ans,quo);
    }
    if(n%s==0) ans=std::min(ans,n/s);
    if(ans==INF) ans=-1;
    cout<<ans<<endl;
}