From Wikipedia:

In modular arithmetic,   the modular multiplicative inverse of an integer    a    modulo    m    is an integer    x    such that





a

x
≡
1


(
mod

m
)

.


{\displaystyle a\,x\equiv 1{\pmod {m}}.}


Or in other words, such that:





∃
k
∈

Z

,

a

x
=
1
+
k

m


{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}


It can be shown that such an inverse exists   if and only if    a    and    m    are coprime,   but we will ignore this for this task.



Task

Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language,   compute the modular inverse of   42 modulo 2017.




sub inverse($n, :$modulo) {
    my ($c, $d, $uc, $vc, $ud, $vd) = ($n % $modulo, $modulo, 1, 0, 0, 1);
    my $q;
    while $c != 0 {
        ($q, $c, $d) = ($d div $c, $d % $c, $c);
        ($uc, $vc, $ud, $vd) = ($ud - $q*$uc, $vd - $q*$vc, $uc, $vc);
    }
    return $ud % $modulo;
}
 
say inverse 42, :modulo(2017)