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: Time Limit Exceeded)

#include <algorithm>
#include <bitset>
#include <cassert>
#include <complex>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <map>
#include <memory>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <typeinfo>
#include <utility>
#include <vector>
#include <array>
#include <chrono>
#include <random>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#define INIT std::ios::sync_with_stdio(false);std::cin.tie(0);
#define VAR(type, ...)type __VA_ARGS__;Scan(__VA_ARGS__);
template<typename T> void Scan(T& t) { std::cin >> t; }
template<typename First, typename...Rest>void Scan(First& first, Rest&...rest) { std::cin >> first; Scan(rest...); }
#define OUT(d) std::cout<<d;
#define FOUT(n, d) std::cout<<std::fixed<<std::setprecision(n)<<d;
#define SOUT(n, c, d) std::cout<<std::setw(n)<<std::setfill(c)<<d;
#define SP std::cout<<" ";
#define TAB std::cout<<"\t";
#define BR std::cout<<"\n";
#define FLUSH std::cout<<std::flush;
#define VEC(type, c, n) std::vector<type> c(n);for(auto& i:c)std::cin>>i;
#define MAT(type, c, m, n) std::vector<std::vector<type>> c(m, std::vector<type>(n));for(auto& r:c)for(auto& i:r)std::cin>>i;
#define ALL(a) (a).begin(),(a).end()
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define RFOR(i, a, b) for(int i=(b)-1;i>=(a);--i)
#define REP(i, n) for(int i=0;i<int(n);++i)
#define RREP(i, n) for(int i=(n)-1;i>=0;--i)
#define FORLL(i, a, b) for(ll i=(a);i<(b);++i)
#define RFORLL(i, a, b) for(ll i=(b)-1;i>=(a);--i)
#define REPLL(i, n) for(ll i=0;i<ll(n);++i)
#define RREPLL(i, n) for(ll i=(n)-1;i>=0;--i)
#define PAIR std::pair<int, int>
#define IN(a, x, b) (a<=x && x<b)
#define SHOW(d) {std::cerr << #d << "\t:" << d << "\n";}
#define SHOWVECTOR(v) {std::cout << #v << "\t:";for(const auto& i : v){std::cout << i << " ";}std::cout << "\n";}
#define SHOWVECTOR2(v) {std::cout << #v << "\t:\n";for(const auto& i : v){for(const auto& j : i){std::cout << j << " ";}std::cout << "\n";}}
#define SHOWPAIRVECTOR2(v) {std::cout << #v << "\t:\n";for(const auto& i : v){for(const auto& j : i){std::cout<<'('<<j.first<<", "<<j.second<<") ";}std::cout << "\n";}}
#define SHOWPAIRVECTOR(v) {for(const auto& i:v){std::cout<<'('<<i.first<<", "<<i.second<<") ";}std::cout<<"\n";}
#define CHECKTIME(state) {auto start=std::chrono::system_clock::now();state();auto end=std::chrono::system_clock::now();auto res=std::chrono::duration_cast<std::chrono::nanoseconds>(end-start).count();std::cerr<<"[Time:"<<res<<"ns  ("<<res/(1.0e9)<<"s)]\n";}
#define SHOWQUEUE(a) {std::queue<decltype(a.front())> tmp(a);std::cout << #a << "\t:";for(int i=0; i<static_cast<int>(a.size()); ++i){std::cout << tmp.front() << "\n";tmp.pop();}std::cout << "\n";}

//#define int ll
using ll = long long;
using ull = unsigned long long;
constexpr int INFINT = 1 << 30;
constexpr ll INFLL = 1LL << 60;
constexpr double EPS = 0.0000000001;
constexpr int MOD = 1000000007;

/*
f(b, n) = s

n := sum{a_k*b^k | 0<=k}
s := sum{a_k | 0<=k}

(1) 2 <= b <= Sqrt(n) (範囲10^6程度)
	-> 全   探   索
(2) Sqrt(n) < b < n
	-> n < b^2
	   ∴ n := pb+q と書ける！ (0 <= p, q < b)
	   このとき s = p + q
	   さらに、
	     n = pb+q
		   > p^2 + q
		   > p^2
	   ∴ 0 < p < Sqrt(n) (範囲10^6程度)
			-> 全   探   索   可   能
	   p=p'のとき、
	     q = n%p'
	   となる
	   
∴
(1)のとき、
ans = min{b | f(b, n) = s (b∊[2, Sqrt(n)]) }

(2)のとき、
ans = min{b | n=pb+q (p∊(0, Sqrt(n)), q = n%p)}
*/

ll f(ll b, ll n) {
	ll res = 0;
	while (n != 0) {
		res += n%b;
		n /= b;
	}
	return res;
}

signed main() {
	INIT;
	VAR(ll, n, s);
	ll ans = INFLL;
	if (n == s) {
		ans = n + 1;
	}
	else {
		FORLL(b, 2, std::sqrt(n) + 1) {
			if (f(b, n) == s) ans = std::min(ans, b);
		}
		FORLL(p, 1, std::sqrt(n)) {
			ll q = s - p;
			ll b = (n - q) / p;
			if (f(b, n) == s) ans = std::min(ans, b);
		}
	}
	OUT(((ans != INFLL) ? ans : -1))BR;
	return 0;
}