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>
#include <stdlib.h>
#include <stdint.h>
#include <math.h>

int64_t test(uint64_t n, uint64_t s);
int main(void) {
    char *p;
    char str[16];
    uint64_t n,s;

    p = fgets(str, 16, stdin);
    n = strtoul(p,0,0);
    p = fgets(str, 16, stdin);
    s = strtoul(p,0,0);

    printf("%ld\n", test(n,s));
//    printf("%ld\n", test(87654UL,30UL));
//    printf("%ld\n", test(87654UL,138UL));
//    printf("%ld\n", test(87654UL,45678UL));
//    printf("%ld\n", test(31415926535UL,1UL));
//    printf("%ld\n", test(1UL,31415926535UL));
//    printf("%ld\n", test(100000000000,1));
    
    return 0;
}

int64_t test(uint64_t n, uint64_t s) {
    uint64_t i,j;
    uint64_t tmp;
    uint64_t sum;
    uint64_t sqrtN;
    uint64_t p;
    int64_t ret = -1;
    if(n == s) {
        return n+1;
    } else if(n < s){
        return -1;
    } else if(s == 1){
        return n;
    }

    sqrtN = (uint64_t)sqrt((double)n) + 1;

    for(i=2UL;i<=sqrtN;i++){
        tmp = n;
        sum = 0;
        while(tmp > 0) {
            sum += (tmp % i);
            tmp /= i;
        }
        if(sum == s){
            ret = i;
            break;
        }
    }
    if(ret == -1){
    for(p=1;p<=sqrtN;p++){
        if((((n-s) % p) == 0)
            && (((n % ((n-s)/p + 1)) + (n-s) + p) == n)){
            //printf("p:%lu\n",p);
            ret = 1 + (n-s)/p;
            break;
        }
   
    }
    }
    
    return ret;
}