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>

using namespace std;
typedef pair<int,int> pii;
typedef long long int ll;

#define INF 1 << 29
#define REP(i,n) for(ll i=0; i<(int)(n); i++)
#define FOR(i,k,n) for(ll i=(k);i<(int)(n);i++)



vector<bool> gen_sosuu(vector<bool> sosuu){
  int size = sosuu.size();
  REP(i,size){
    sosuu[i] = true;
  }
  sosuu[0] = false;
  sosuu[1] = false;
  sosuu[2] = true;
  FOR(i,2,sqrt(size)+1){
    if(sosuu[i] == false) continue;
    for(int j = 2; i*j<size;j++){
      sosuu[i*j] =false;
    }
  }
  return sosuu;
}

ll n, s;
ll b = LLONG_MAX;

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


int main(){
  cin >> n >> s;
  if(s==n){
    cout << n+1 << endl;
    return 0;
  }
  FOR(i,2,sqrt(n)+1){
    if(f(i,n) == s){
      cout << i <<endl;
      return 0;
    }
  }
  ll ans = LLONG_MAX;
  FOR(i,1,sqrt(n)+1){
    ll b = (n-s)/i + 1;
    if(f(b,n) == 1){
      ans = max(b,ans);
    }
  }
  if(ans == LLONG_MAX){
    cout << -1 << endl;
    return 0;
  }
  cout << b<<endl;
}