Median Weight Bead

There are $N$ beads of the same shape and size but with different weights. $N$ is odd, and the beads are labeled $1, 2, \dots, N$. Your task is to identify the bead whose weight is the median — the $\frac{N+1}{2}$-th lightest bead among all $N$ beads.

A scale lets us compare any two beads and decide which is heavier. After $M$ comparisons we know that some beads are heavier than others, and this relation is transitive: if bead $A$ is heavier than bead $B$ and bead $B$ is heavier than bead $C$, then $A$ is heavier than $C$. Using only this information, we want to discard every bead that can never be the median.

A bead can never be the median if at least $\frac{N+1}{2}$ other beads are known to be heavier than it, or at least $\frac{N+1}{2}$ other beads are known to be lighter than it.

For example, suppose $N = 5$ with the following $M = 4$ results:

1. Bead 2 is heavier than Bead 1.
2. Bead 4 is heavier than Bead 3.
3. Bead 5 is heavier than Bead 1.
4. Bead 4 is heavier than Bead 2.

We still cannot tell exactly which bead is the median, but Bead 1 and Bead 4 can never be it: Beads 2, 4, and 5 are all heavier than Bead 1, and Beads 1, 2, and 3 are all lighter than Bead 4. Since $\frac{N+1}{2} = 3$, both beads meet the removal condition.

Write a program that counts how many beads can never be the median.

The first line contains an integer $t$ ($1 \le t \le 11$), the number of test cases. The data for each test case follows.

The first line of each test case contains two integers $N$ ($1 \le N \le 99$) and $M$, where $N$ is the number of beads and $M$ is the number of comparisons. Each of the next $M$ lines contains two integers $a$ and $b$, meaning that bead $a$ is heavier than bead $b$.

For each test case, print a single line containing the number of beads that can never be the median.