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++ (GCC 9.2.1) (judge verdict: Accepted)

/*#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll f (ll b , ll n){
    if (n < b) return n;
    return f(b , n / b) + (n % b);
}
int main()
{
    ios::sync_with_stdio(false);
    ll n , s; cin >> n >> s;
    if (n == s) {
        cout << n + 1 << endl;
        return 0;
    }
    ll ans = -1;
    for (ll i = 2 ; i * i <= n ; i++){
        if (f(i , n) == s) {
            ans = i;
            break;
        }
    }
    if (ans != -1){
        cout << ans << endl;
        return 0;
    }
    for (ll i = 1 ; i * i <= n - s ; i++){
        if (n % i == 0){
            ll b = (n - s) / i + 1;
            if (b <= n / b) continue;
            if (b < 2) continue;
            ll k = n / b , r = n % b;
            if (s != r + k) continue;
            if (ans == -1 || ans > b) ans = b;
        }
    }
    cout << ans << endl;
    return 0;
}*/

#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll n,s;
bool check(ll x){
	if(n/x>=x)return 0;
	if(x<2)return 0;
	ll k=n/x;ll r=n%x;
	if(s==k+r)return 1;
	return 0;
}
ll js(ll x,ll b){
	ll ans=0;
	while(x){
		ans+=x%b;
		x/=b;
	}
	return ans;
}
int main()
{
	//freopen("H:\\c++1\\in.txt","r",stdin);
	//freopen("H:\\c++1\\out.txt","w",stdout);
	cin>>n>>s;
	ll b=1e18;
	if(n==s)b=n+1;
	for(ll t=2;t * t <= n;t++){
		ll cnt=js(n,t);
		if(cnt==s){
			b=min(b,t);
			break;
		}
	}
	if (b != 1e18){
        cout << b << endl;
        return 0;
	}
	if(n>s){
		ll temp=n-s;
		for(ll t=1;t<=sqrt(temp);t++){
			if(temp%t==0){
				ll x=t+1,y=temp/t+1;
				if(check(y))b=min(b,y);
			}
		}
	}
	if(b==1e18)printf("-1\n");
	else printf("%lld\n",b);
	return 0;
}


/*
82914867733
1676425945
*/