Consider the following exercise, found in a generic linear
algebra textbook.
Let $A$ be an
$n \times n$ matrix.
Prove that the following statements are equivalent:

$A$ is
invertible.

$Ax = b$ has
exactly one solution for every $n \times 1$ matrix $b$.

$Ax = b$ is
consistent for every $n
\times 1$ matrix $b$.

$Ax = 0$ has only
the trivial solution $x =
0$.
The typical way to solve such an exercise is to show a
series of implications. For instance, one can proceed by
showing that (a) implies (b), that (b) implies (c), that (c)
implies (d), and finally that (d) implies (a). These four
implications show that the four statements are equivalent.
Another way would be to show that (a) is equivalent to (b)
(by proving that (a) implies (b) and that (b) implies (a)),
that (b) is equivalent to (c), and that (c) is equivalent to
(d). However, this way requires proving six implications, which
is clearly a lot more work than just proving four
implications!
I have been given some similar tasks, and have already
started proving some implications. Now I wonder, how many more
implications do I have to prove? Can you help me determine
this?
Input
On the first line one positive number: the number of
testcases, at most 100. After that per testcase:

One line containing two integers $n$ ($1 \le n \le 20\, 000$) and
$m$ $(0 \le m \le 50\, 000)$: the
number of statements and the number of implications that
have already been proved.

$m$ lines with two
integers $s_1$ and
$s_2$ ($1\le s_1,s_2\le n$ and
$s_1 \ne s_2$) each,
indicating that it has been proved that statement
$s_1$ implies
statement $s_2$.
Output
Per testcase:
Sample Input 1 
Sample Output 1 
2
4 0
3 2
1 2
1 3

4
2
