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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
typedef long long ll;

ll min(ll a,ll b){return a<b?a:b;}

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

int main(void){
    ll n,s,ans;
    scanf("%lld%lld",&n,&s);
    if (s>n){
        ans=-1;
    } else if (n==s){
        ans=n+1;
    } else {
        for (ll i=2; i<=316227; i++){
            if (DigitSum(i,n)==s){
                ans=i;
                break;
            }
        }
        for (ll i=1; i<=316227; i++){
            ll b=(n-s)/i+1;
            if ((n-s)%i==0 && DigitSum(b,n)==s){
                ans=b;
                break;
            }
        }
    }
    printf("%lld\n",ans);
	return 0 ;
}