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)

#include <bits/stdc++.h>
using namespace std;
#define int long long   // <-----!!!!!!!!!!!!!!!!!!!

#define rep(i,n) for (int i=0;i<(n);i++)
#define rep2(i,a,b) for (int i=(a);i<(b);i++)
#define rrep(i,n) for (int i=(n)-1;i>=0;i--)
#define rrep2(i,a,b) for (int i=(b)-1;i>=(a);i--)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define printV(v) for(auto&& x : v){cout << x << " ";} cout << endl
#define printVV(vv) for(auto&& v : vv){for(auto&& x : v){cout << x << " ";}cout << endl;}
#define printP(p) cout << p.first << " " << p.second << endl
#define printVP(vp) for(auto&& p : vp) printP(p);

typedef long long ll;
typedef pair<int, int> Pii;
typedef tuple<int, int, int> TUPLE;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;
typedef vector<Pii> vp;
// const int inf = 1e9;
const int inf = 1e18;
const int mod = 1e9 + 7;
typedef vector<vector<int>> Graph;

int f(int b, int n) {
    if (n < b) return n;
    return f(b, n / b) + n % b;
}

signed main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);

    int n, s;
    cin >> n >> s;

    if (n == s) {
        cout << n + 1ll << endl;
        return 0;
    }

    // b <= sqrt(n) は全探索
    for (int b = 2ll; b * b <= n; b++) {
        if (f(b, n) == s) {
            cout << b << endl;
            return 0;
        }
    }

    // b > sqrt(n)のとき
    // nはb進数で高々2桁なので、
    // n = pb+q
    // s = p+q
    // -> n-s = p(b-1)
    // ∴ bはn-sの約数+1
    int ans = inf;
    for (int d = 1ll; d * d <= n - s; d++) {
        if ((n - s) % d == 0ll) {
            int b = d + 1ll;
            if (f(b, n) == s) {
                ans = min(ans, b);
            }
            b = (n - s) / d + 1ll;
            if (f(b, n) == s) {
                ans = min(ans, b);
            }
        }
    }

    cout << (ans == inf ? -1 : ans) << endl;
}