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 <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cstring>
#include <cctype>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <map>
#include <list>
#include <stdio.h>
#include <string.h>
#include <cstdlib>
#include <math.h>
#include <bitset>
#include <iterator>
#include <iomanip>
#include <sstream>
#include <numeric>
#define INF 200000000000
using namespace std;
int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};
long long gcd(long long a,long long b){if(a%b==0){return b;}else return gcd(b,a%b);}
long long lcm(long long a,long long b){if(a==0){return b;} return a/gcd(a,b)*b;}
#define MOD 1000000007
#define sym cout<<"---------"<<endl;
#define ll long long
#define PL pair<long long, long long>
#define P pair<int,int>
#define mk make_pair
#define en endl
#define WHITE 1
#define BLACK 2
#define GRAY 1
#define RE return 0
#define int ll

int solve(int num,int mod){
  int sum=0;
  while(num){
    int dig=num%mod;
    sum+=dig;
    num=floor(num/mod);
  }
  return sum;
}

signed main(){
  int n,s; cin>>n>>s;
  
  if(s==1){
    cout<<n<<en;
    RE;
  }else if(n<s){
    cout<<-1<<en; RE;
  }
  for(int i=2; i*i<=n; i++){
    int ret=solve(n, i);
    if(ret==s){
      cout<<i<<en;
      RE;
    }
  }
  int ans=INF;
  for(int i=1; i*i<n; i++){
    if((n-s)%i) continue;
    if(solve(n, (n-s)/i+1)==s){
      ans=min(ans,(n-s)/i+1);
    }
  }
  if(ans==INF) cout<<-1<<en;
  else cout<<ans<<en;
}