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)

//#pragma comment(linker, "/STACK:102400000,102400000")
#include <fstream>
#include <iostream>
#include <functional>
#include <algorithm>
#include <cstring>
#include <vector>
#include <string>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <deque>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;

typedef long long ll;
#define eps 1e-10
#define LL_INF 0x33ffffffffffffff
#define INF 0x3f3f3f3f
#define mem(a, b) memset(a, b, sizeof(a))
#define pper(i,n,m) for(int i = n;i >= m; i--)
#define repp(i, n, m) for (int i = n; i <= m; i++)
#define rep(i, n, m) for (int i = n; i < m; i++)
#define sa(n) scanf("%d", &(n))
#define mp make_pair
#define ff first
#define ss second
#define pb push_back

const int maxn = 505;
const double PI = acos(-1.0);
const ll mod = 1e6 + 3;
ll po(ll a, ll b,ll mod) { ll res = 1; a %= mod; for (; b; b >>= 1) { if (b & 1)res = res*a%mod; a = a*a%mod; }return res; }
ll gcd(ll a,ll b){ if(a==0){return b;}else{ return gcd(b%a,a);}}

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

void solve()
{
	scanf("%lld%lld",&n,&s);
	ll up = sqrt(1.0*n)+1;
	for(int i=2;i<=up;i++)
	{
		if(f(i,n)==s)
		{
			printf("%d",i);
			return;
		}
	}
	ll t = n-s;
	if(t < 0)
	{
		puts("-1");
		return;
	}
	if(t==0)
	{
		printf("%lld",n+1);
		return;
	}
	for(int i=1;i<=up;i++)
	{
		if(t%i==0)
		{
			ll b = t/i+1;
			ll p = i;
			ll q = s-p;
			if(q >= b)continue;
			if(p >= b)continue;
			//cout<<p<<" "<<q<<endl;
			if(p*b+q == n)
			{
				printf("%lld",b);
				return;
			}
		}
	}
	puts("-1");
}

int main()
{

    solve();
	return 0;
}