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: Accepted)

// Author: Vamsi Krishna Reddy Satti
// With love for Competitive Programming!
 
#include <bits/stdc++.h>
using namespace std;
 
#define fast_io ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
#define cout_precision cout.setf(ios::fixed); cout.precision(32);
 
using ll = int64_t; using vll = vector<ll>; using vvll = vector<vll>;
using pll = pair<ll, ll>; using vpll = vector<pll>; using vvpll = vector<vpll>;
using ld = long double;
 
// ------------------------------------------------------------------------------------------------

ll n, s, d, ans;

ll valid(ll m) {
    ll sum = 0;
    for (ll x = n; x > 0; x /= m) {
        sum += x % m;
    }
    return sum == s;
}

int main() {
    fast_io; cout_precision;
    cin >> n >> s;
    if (n < s) {
        cout << -1 << endl;
        return 0;
    } else if (n == s) {
        cout << n + 1 << endl;
        return 0;
    }
    d = n - s, ans = INT64_MAX;
    for (ll i = 1; i * i <= d; i++) {
        if (d % i != 0) {
            continue;
        }
        if (valid(i + 1)) {
            ans = min(ans, i + 1);
        }
        if (valid(d / i + 1)) {
            ans = min(ans, d / i + 1);
        }
    }
    cout << ((ans == INT64_MAX) ? -1 : ans) << endl;
}