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 Go (1.6) (judge verdict: Wrong Answer)

package main

import "fmt"

func digit_sum(x, b int64) int64 {
    ret := int64(0)
    for x > 0 {
        ret += x % b
        x /= b
    }
    return ret
}

func solve(n, s int64) int64 {
    if s > n {
        return -1
    }
    if s == n {
        return n + 1
    }
    for b := int64(2); b * b <= n; b++ {
        if digit_sum(n, b) == s {
            return b
        }
    }
    for p := int64(1); p * p < n; p++ {
        b := (n - s) / p + 1
        fmt.Println(b)
        if digit_sum(n, b) == s {
            return b
        }
    }
    return -1
}

func main() {
    var n, s int64
    fmt.Scan(&n, &s)
    fmt.Println(solve(n, s))
}