E. Tree or not Tree

гаралт стандарт гаралт

You are given an undirected connected graph $G$ consisting of $n$ vertexes and $n$ edges. $G$ contains no self-loops or multiple edges. Let each edge has two states: on and off. Initially all edges are switched off.

You are also given $m$ queries represented as $(v, u)$ -- change the state of all edges on the shortest path from vertex $v$ to vertex $u$ in graph $G$. If there are several such paths, the lexicographically minimal one is chosen. More formally, let us consider all shortest paths from vertex $v$ to vertex $u$ as the sequences of vertexes $v, v_{1}, v_{2}, ..., u$. Among such sequences we choose the lexicographically minimal one.

After each query you should tell how many connected components has the graph whose vertexes coincide with the vertexes of graph $G$ and edges coincide with the switched on edges of graph $G$.

Оролт

The first line contains two integers $n$ and $m$ ($3 ≤ n ≤ 10^{5}$, $1 ≤ m ≤ 10^{5}$). Then $n$ lines describe the graph edges as $a$ $b$ ($1 ≤ a, b ≤ n$). Next $m$ lines contain the queries as $v$ $u$ ($1 ≤ v, u ≤ n$).

It is guaranteed that the graph is connected, does not have any self-loops or multiple edges.

Гаралт

Print $m$ lines, each containing one integer -- the query results.

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Жишээ тэстүүд

Оролт
5 2
2 1
4 3
2 4
2 5
4 1
5 4
1 5

Гаралт
3
3

Оролт
6 2
4 6
4 3
1 2
6 5
1 5
1 4
2 5
2 6

Гаралт
4
3


Тэмдэглэл

Let's consider the first sample. We'll highlight the switched on edges blue on the image.

• The graph before applying any operations. No graph edges are switched on, that's why there initially are 5 connected components.

* The graph after query $v = 5, u = 4$. We can see that the graph has three components if we only consider the switched on edges.

* The graph after query $v = 1, u = 5$. We can see that the graph has three components if we only consider the switched on edges.

Lexicographical comparison of two sequences of equal length of $k$ numbers should be done as follows. Sequence $x$ is lexicographically less than sequence $y$ if exists such $i$ ($1 ≤ i ≤ k$), so that $x_{i} < y_{i}$, and for any $j$ ($1 ≤ j < i$) $x_{j} = y_{j}$.

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