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)

/*    三玖天下第一！！！  */
#include<bits/stdc++.h>
using namespace std;
const long long inf=0x3f3f3f3f3f3f3f3f;long long n,s,ans=inf;
long long f(long long bb,long long nn){return nn<bb?nn:(f(bb,nn/bb)+nn%bb);}
inline void in(){scanf("%lld%lld",&n,&s);}
inline void work()
{
	if(n<s) {puts("-1");return;}else if(n==s) {printf("%lld\n",n+1);return;}
	for(long long i=2;i*i<=n;i++) if(f(i,n)==s) {printf("%lld\n",i);return;} 
	for(long long i=1,temp;i*i<=(n-s);i--) if((n-s)%i==0) {temp=((n-s)/i)+1;if(f(temp,n)==s) ans=min(ans,temp);}
	printf("%lld\n",ans==inf?-1:ans);
}
int main(){in();work();;return 0;}