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>
#define base 1000000007LL
#define ll long long
#define X first
#define Y second
#define mp make_pair
#define pb push_back
#define CLR(a) memset(a,0,sizeof(a))
#define FOR(i,a,b) for(ll i=(a),_b=(b); i<=_b; i++)
#define FORE(i,a,b) for(ll i=(a),_b=(b); i>=_b; i--)

using namespace std;

typedef pair<int, int> II;
typedef vector<II> vi;

ll n, s;

int main()
{
    ios::sync_with_stdio(0);
    //freopen("inp.txt", "r", stdin);
    cin >> n >> s;
    if (n < s) {
        cout << -1;
        return 0;
    }
    if (n == s) {
        cout << n+1;
        return 0;
    }
    ll res = 1000000000000LL;
    ll o = n-s;
    FOR(i,1,(ll)sqrt(o))
        if (o % i == 0) {
            ll k = o / i;
            if (i+1 <= n && i+1 > (ll)sqrt(n) && n / (i+1) == k) res = min(res, i+1);
            if (k+1 <= n && k+1 > (ll)sqrt(n) && n / (k+1) == i) res = min(res, k+1);
        }
    FOR(i,2,(ll)sqrt(n)) {
        ll p = n;
        ll r = 0;
        while (p) {
            r += p % i;
            p /= i;
        }
        if (r == s) {
            cout << i;
            return 0;
        }
    }
    if (res == 1000000000000LL) res = -1;
    cout << res;
    return 0;
}