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: Runtime Error)

#include<bits/stdc++.h>
#define fi first
#define se second
#define show(x) cout<<#x<<"="<<x<<"\n"
typedef long long ll;
template<typename T>
void print_to(std::ostream &os,const char *,const char *tail,const T &fst){
  os<<fst<<tail;
}
template<typename Fst,typename... Rst>
void print_to(std::ostream &os,const char *del,const char *tail,const Fst &fst,const Rst &... rst){
  os<<fst<<del;
  print_to(os,del,tail,rst...);
}
template<typename Iter>
void print_to_(std::ostream &os,const char *del,const char *tail,Iter bgn,Iter end){
  for(Iter it=bgn;it!=end;){
    os<<*it;
    if(++it!=end)os<<del;
    else os<<tail;
  }
}
template<typename Fst,typename... Rst>
void println(const Fst &fst,const Rst &... rst){
  print_to(std::cout,"\n","\n",fst,rst...);
}
template<typename Fst,typename... Rst>
void print(const Fst &fst,const Rst &... rst){
  print_to(std::cout," ","\n",fst,rst...);
}
template<typename Iter>
void println_(Iter bgn,Iter end){
  print_to_(std::cout," ","\n",bgn,end);
}
using namespace std;
const ll inf=(ll)1e14;
//const int MOD=(ll)1e10+7;
const int dy[]={1,0,-1};
const int dx[]={1,0,-1};
ll n,m,h,w;
string s;

ll cal(ll b,ll n){
  if(n<b)return n;
  else return cal(b,n/b)+n%b;
}

int main(){
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(10);
  cout<<fixed;
#ifdef LOCAL_DEFINE
  FILE *stream1;
  //FILE *stream2;
  stream1=freopen("in","r",stdin);
  //stream2=freopen("out","w",stdout);
  if(stream1==NULL)return 0;
  //if(stream2==NULL)return 0;
#endif
  cin>>n>>m;
  if(n==m){
    print(n+1);
    return 0;
  }else if(m==1){
    print(n);
    return 0;
  }else if(m>n){
    print(-1);
    return 0;
  }
  for(int i=2;i<=(int)sqrt(n)+1;i++){
    if(cal(i,n)==m){
      print(i);
      return 0;
    }
  }
  ll minans=inf;
  for(int i=1;i<=(int)sqrt(n)+1;i++){
    ll q=i;
    if(cal((n-m-q)/q,n)==m){
      minans=min(minans,(n-m-q)/q);
    }
  }
  if(minans==inf)print(-1);
  else print(minans);
#ifdef LOCAL_DEFINE
  cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<"s.\n";
  fclose(stream1);
  //fclose(stream2);
#endif
  return 0;
}