#S0105. Babysitting
Babysitting
Description
Theofanis wants to play video games, however he should also take care of his sister. Since Theofanis is a CS major, he found a way to do both. He will install some cameras in his house in order to make sure his sister is okay.
His house is an undirected graph with $n$ nodes and $m$ edges. His sister likes to play at the edges of the graph, so he has to install a camera to at least one endpoint of every edge of the graph. Theofanis wants to find a vertex cover that maximizes the minimum difference between indices of the chosen nodes.
More formally, let $a_1, a_2, \ldots, a_k$ be a vertex cover of the graph. Let the minimum difference between indices of the chosen nodes be the minimum $\lvert a_i - a_j \rvert$ (where $i \neq j$) out of the nodes that you chose. If $k = 1$ then we assume that the minimum difference between indices of the chosen nodes is $n$.
Can you find the maximum possible minimum difference between indices of the chosen nodes over all vertex covers?
Input
The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.
The first line of each test case contains two integers $n$ and $m$ ($1 \le n \le 10^{5}, 1 \le m \le 2 \cdot 10^{5}$) — the number of nodes and the number of edges.
The $i$-th of the following $m$ lines in the test case contains two positive integers $u_i$ and $v_i$ ($1 \le u_i,v_i \le n$), meaning that there exists an edge between them in the graph.
It is guaranteed that the sum of $n$ over all test cases does not exceed $10^{5}$.
It is guaranteed that the sum of $m$ over all test cases does not exceed $2 \cdot 10^{5}$.
Output
For each test case, print the maximum minimum difference between indices of the chosen nodes over all vertex covers.
3
7 6
1 2
1 3
1 4
1 6
2 3
5 7
3 3
1 2
1 3
1 1
2 4
1 2
1 2
2 1
1 1
2
3
2
Note
In the first test case, we can install cameras at nodes $1$, $3$, and $7$, so the answer is $2$.
![](./315/file/AxczGVwG.png)
In the second test case, we can install only one camera at node $1$, so the answer is $3$.