I am having some repeated trouble getting the correct answer on linear congruences. Consider the following
$$12x \equiv 1 \pmod {77} $$
$12$ and $77$ are relatively prime so this congruence has a solution. We search for a linear combination of $12$ and $77$ using the extended Euclidean algorithm.
$$77=12(6)+5\\ 12=5(2)+2\\ 5=2(2)+1$$
We now solve for the remainders
$$1=5-2(2)\\ 2=12-5(2)\\ 5=77-12(6)$$
Back substituting we find
$1=77(5)+12(-32)$
The solution to this congruence is $45$ and $-32 \equiv 45$ (mod 77). What am I failing to do properly as to get the first positive solution?