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: Runtime Error)

#include <iostream>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <cmath>
#include <iomanip>
#include <queue>

using namespace std;

#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define REP(i, n) FOR(i, 0, n)
#define SORT(c) sort((c).begin(), (c).end())

typedef long long ll;
typedef pair<int, int> P;
typedef vector<int> V;
typedef map<int, int> M;

const ll INF = 1e18;
const ll MOD = 1e9 + 7;

ll sum(ll b, ll n)
{
    if (b > n)
        return n;

    return sum(b, n / b) + n % b;
}

int main()
{
    ll n, s, ans;
    cin >> n >> s;

    ans = INF;

    ll sq = sqrt(n);
    if (sq * sq == n)
        sq--;

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

    for (ll i = 2; i <= sqrt(n); i++)
    {
        if (sum(i, n) == s)
        {
            ans = min(i, ans);
        }
    }

    for (ll i = 1; i <= sq; i++)
    {
        ll b = (n - s) / i + 1;
        if (b == 0)
            continue;
        if (sum(b, n) == s)
        {
            ans = min(b, ans);
        }
    }

    if (ans == INF)
        cout << -1 << endl;
    else
        cout << ans << endl;

    return 0;
}