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 SPEED ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define fileio freopen("in.in", "r", stdin),freopen("out.out", "w", stdout);
#define ll long long int
#define FF first
#define SS second
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define pll pair<long long int,long long int>
#define sd(x) scanf("%d",&x)
#define slld(x) scanf("%lld",&x)
#define pd(x) printf("%d\n",x)
#define plld(x) printf("%lld\n",x)
#define pss printf
#define MOD 1000000007
#define INF 1e18
#define eps 0.00001
#define endl '\n'
#define debug(n1) cout<<n1<<endl
ll n,s;
inline bool check(ll x,ll i)
{
	if(i==1)
		return 0;
	if(i==0)return 0;
	ll an=0;
	while(x)
	{
		an+=x%i;
		x/=i;
	}
	return an==s;
}
int main()
{
	SPEED;
	cin>>n>>s;
	for(int i=2;i<=1e6;i++)
		if(check(n,i))
			return cout<<i<<endl,0;
	for(int i=1e6;i>=1;i--)
	{
		if(check(n,n/i))
			return cout<<n/i<<endl,0;
	}
	cout<<-1<<endl;
	return 0;	
}