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 PHP7 (7.0.15) (judge verdict: Wrong Answer)

<?php
$n = intval(fgets(STDIN));
$s = intval(fgets(STDIN));

if($n == $s){
    echo $n+1,"\n";exit;
}
//$bがsqrt($n)より小さい場合
for($i = 2, $I = sqrt($n); $i < $I; $i++){
    $m = $n;
    $sum = 0;
    while($m){
        $sum += $m % $i;
        $m = intdiv($m, $i);
    }
    if($sum == $s){
        echo $i,"\n";exit;
    }
}
//$bがsqrt($n)以上の場合　上位桁を決め打つと下位の桁は決まる
$minB = 10**18;
for($p = 1, $P = sqrt($n); $p <= $P; $p++){
    $q = $s - $p;
    if(($n-$q) % $p != 0)continue;
    $b = ($n-$q)/$p;
    if($p >= $b)continue;
    if($q >= $b)continue;
    if($q < 0)continue;
    $minB = min($minB, $b);
}
if($minB == 10**18){
    echo -1;
}else{
    printf("%d", $minB);
}