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 <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <cmath>
#include <deque>
#include <iterator>
#include <map>
#include <queue>
#include <string>
#include <tuple>
#include <utility>
#include <limits>
#include <iomanip>
using namespace std;

using ll=long long;
template<class T> using V = vector<T>;
template<class T, class U> using P = pair<T, U>;
using vll = V<ll>;
using vvll = V<vll>;
#define rep(i, k, n) for (ll i=k; i<(ll)n; ++i)
#define REP(i, n) rep(i, 0, n)
template<class T> inline bool chmax(T& a, T b) {if (a<b) {a=b; return true;} return false;}
template<class T> inline bool chmin(T& a, T b) {if (a>b) {a=b; return true;} return false;}

const ll MOD = 1000000007;
const ll HIGHINF = (ll)1e18;

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

int main() {
  ll n, s; cin >> n >> s;
  ll b = 2;
  if (n == s) {cout << n+1 << endl; return 0;}
  while (b*b <= n) {
    if (f(b, n) == s) {cout << b << endl; return 0;}
    b++;
  }
  ll p = 1LL;
  ll minb = HIGHINF;
  while (p*p < n) {
    b = (n-s+p) / p;
    if (b < 2) {p++; continue;}
    if (f(b, n) == s) chmin(minb, b);
    p++;
  }
  if (minb == HIGHINF) cout << -1 << endl;
  else cout << minb << endl;
  return 0;
}