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: Wrong Answer)

// ConsoleApplication9.cpp : アプリケーションのエントリ ポイントを定義します。
//


#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<math.h>
#include<iomanip>
#include<set>
#include<numeric>
#include<cstring>
#include<cstdio>
#include<functional>

#define REP(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORR(i, m, n) for(int i = m;i >= n;i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define REVERSE(v,n) reverse(v,v+n);
#define VREVERSE(v) reverse(v.begin(), v.end());
#define ll long long
#define pb(a) push_back(a)
#define INF 9999999999
#define m0(x) memset(x,0,sizeof(x))
#define fill(x,y) memset(x,y,sizeof(x))
#define print(x) cout<<x<<endl;
#define pe(x) cout<<x<<" ";
#define lb(v,n) lower_bound(v.begin(), v.end(), n);
//#define int long long


using namespace std;


int dy[4] = { 0,0,1,-1 };
int dx[4] = { 1,-1,0,0 };
int dxx[9] = {0, 0,0,1,1,1,-1,-1,-1 };
int dyy[9] = {0, 1,-1,0,1,-1,0,1,-1 };




ll gcd(ll x, ll y) {
	ll m = max(x, y), n = min(x, y);
	if (m%n == 0)return n;
	else return gcd(m%n, n);
}
ll lcm(ll x, ll y) {
	return x / gcd(x, y)*y;
}

ll myPow(ll x, ll n, ll m) {
	if (n == 0)
		return 1;
	if (n % 2 == 0)
		return myPow(x * x % m, n / 2, m);
	else
		return x * myPow(x, n - 1, m) % m;
}

ll pow2(ll a, ll n) {//aのn乗を計算します。
	ll x = 1;
	while (n > 0) {//全てのbitが捨てられるまで。
		if (n & 1) {//1番右のbitが1のとき。
			x = x * a;
		}
		a = a * a;
		n >>= 1;//bit全体を右に1つシフトして一番右を捨てる。
	}
	return x;
}

long long nCr(int n, int r) {
	if (r > n / 2) r = n - r; // because C(n, r) == C(n, n - r)
	long long ans = 1;
	int i;

	for (i = 1; i <= r; i++) {
		ans *= n - r + i;
		ans /= i;
	}

	return ans;
}

const int MAX = 510000;
const int MOD = 1000000007;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < MAX; i++) {
		fac[i] = fac[i - 1] * i % MOD;
		inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
		finv[i] = finv[i - 1] * inv[i] % MOD;
	}
}

// 二項係数計算
long long COM(int n, int k) {
	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

bool arr[100000100];
vector<ll>sosuu;
void Eratosthenes() {
	ll N = 10000010;
	int c = 0;
	for (int i = 0; i < N; i++) {
		arr[i] = 1;
	}
	for (ll i = 2; i < sqrt(N); i++) {
		if (arr[i]) {
			for (ll j = 0; i * (j + 2) < N; j++) {
				arr[i *(j + 2)] = 0;
			}
		}
	}

	for (ll i = 2; i < N; i++) {
		if (arr[i]) {
			sosuu.pb(i);
			//cout << sosuu[c] << " ";
			c++;
		}
	}
	//cout << endl;
	//cout << c << endl;
}

ll stoL(string s) {
	ll n = s.length();
	ll ans = 0;
	for (int i = 0; i < n; i++) {
		ans += pow2(10, n - i - 1)*(ll)(s[i] - '0');
	}
	return ans;
}
ll func(ll b, ll n) {
	if (n < b)return n;
	else return func(b, n / b) + n % b;
}
int main() {
	ll n, s; cin >> n >> s;
	if (s == 1) {
		print(n);
		return 0;
	}
	for (ll b = 2; b*b <= n; b++) {
		if (func(b, n) == s) {
			print(b);
			return 0;
		}
	}

	ll ans = 10e12;
	ll x = n - s;
	for (int i = 2; i*i <= x; i++) {
		if (x%i == 0) {
			ll b = x / i + 1;
			if (func(b, n) == s) {
				ans = min(ans, b);
			}
		}
	}
	if (ans == 10e12) {
		print(-1);
	}
	else print(ans);
	/*
	while (abs(ok - ng) > 1) {
		ll mid = (ok + ng) / 2;
		ll res = func(mid, n);
		pe(mid); print(res);
		if (res >= s) {
			ok = mid;
		}
		else {
			ng = mid;
		}
	}
	if (func(ok, n) == s) {
		print(ok);
	}
	else print(-1);
	*/
}