CF1044F.DFS

Let $T$ be a tree on $n$ vertices. Consider a graph $G_0$, initially equal to $T$. You are given a sequence of $q$ updates, where the $i$-th update is given as a pair of two distinct integers $u_i$ and $v_i$.

For every $i$ from $1$ to $q$, we define the graph $G_i$ as follows:

• If $G_{i-1}$ contains an edge ${u_i, v_i}$, then remove this edge to form $G_i$.
• Otherwise, add this edge to $G_{i-1}$ to form $G_i$.

Formally, $G_i := G_{i-1} \triangle {{u_i, v_i}}$ where $\triangle$ denotes the set symmetric difference.

Furthermore, it is guaranteed that $T$ is always a subgraph of $G_i$. In other words, an update never removes an edge of $T$.

Consider a connected graph $H$ and run a depth-first search on it. One can see that the tree edges (i.e. the edges leading to a not yet visited vertex at the time of traversal) form a spanning tree of the graph $H$. This spanning tree is not generally fixed for a particular graph — it depends on the starting vertex, and on the order in which the neighbors of each vertex are traversed.

We call vertex $w$ good if one can order the neighbors of each vertex in such a way that the depth-first search started from $w$ produces $T$ as the spanning tree. For every $i$ from $1$ to $q$, find and report the number of good vertices.

设 $T$ 是一个包含 $n$ 个顶点的树。考虑图 $G_0$，其初始值等于 $T$。给定一个长度为 $q$ 的更新序列，其中第 $i$ 个更新以一对互异的整数 $u_i$ 和 $v_i$ 给出。

对每个从 $1$ 到 $q$ 的 $i$，我们按如下方式定义图 $G_i$：

• 若 $G_{i-1}$ 中已存在边 $\{u_i, v_i\}$，则将该边从 $G_{i-1}$ 中删除，得到 $G_i$；
• 否则，将该边加入 $G_{i-1}$，得到 $G_i$。

形式化地，$G_i := G_{i-1} \triangle \{u_i, v_i\}$，其中 $\triangle$ 表示集合的对称差。

此外，题目保证 $T$ 始终是 $G_i$ 的子图。换言之，任意一次更新均不会删除 $T$ 中的边。

考虑一个连通图 $H$ 并在其上执行深度优先搜索（DFS）。可以观察到：树边（即在遍历过程中通向尚未访问过的顶点的边）构成图 $H$ 的一棵生成树。该生成树通常并非由图 $H$ 唯一确定——它依赖于 DFS 的起始顶点，以及每个顶点的邻居被遍历的顺序。

若存在一种方式，使得对每个顶点的邻居指定某种遍历顺序，并从顶点 $w$ 出发执行 DFS，所得生成树恰好为 $T$，则称顶点 $w$ 是好的（good）。

对每个从 $1$ 到 $q$ 的 $i$，求出并报告此时“好”顶点的个数。

The first line contains two integers $n$ and $q$ ($3 \le n \le 2\cdot 10^5$, $1 \le q \le 2 \cdot 10^5$) — the number of nodes and the number of updates, respectively.

Each of the next $n-1$ lines contains two integers $u$ and $v$ ($1 \le u, v \le n$, $u \ne v$) — vertices connected by an edge in $T$. It is guaranteed that this graph is a tree.

Each of the next $q$ lines contains two integers $u$ and $v$ ($1 \le u, v \le n$, $u \ne v$) — the endpoints of the edge that is added or removed. It is guaranteed that this edge does not belong to $T$.

第一行包含两个整数 $n$ 和 $q$（$3 \le n \le 2\cdot 10^5$，$1 \le q \le 2 \cdot 10^5$）—— 分别表示节点数量和更新操作次数。

接下来的 $n-1$ 行，每行包含两个整数 $u$ 和 $v$（$1 \le u, v \le n$，$u \ne v$）—— 表示树 $T$ 中由一条边连接的两个顶点。保证该图是一棵树。

再接下来的 $q$ 行，每行包含两个整数 $u$ 和 $v$（$1 \le u, v \le n$，$u \ne v$）—— 表示被添加或删除的边的两个端点。保证该边不属于 $T$。

For each update, print one integer $k$ — the number of good vertices $w$ after the corresponding update.

每次更新后，输出一个整数 $k$ —— 表示对应更新后好顶点 $w$ 的数量。

The first sample is depicted in the following figure.

After the first update, $G$ contains all three possible edges. The result of a DFS is as follows:

• Let the starting vertex be $1$. We have two choices of ordering the neighbors of $1$, either $[2, 3]$ or $[3, 2]$.
  • If we choose the former, then we reach vertex $2$. Regardless of the ordering of its neighbors, the next visited vertex will be $3$. Thus, the spanning tree generated by this DFS will contain edges ${1, 2}$ and ${2, 3}$, which does not equal to $T$.
  • If we choose the latter, we obtain a spanning tree with edges ${1, 3}$ and ${2, 3}$.Hence, there is no way of ordering the neighbors of vertices such that the DFS produces $T$, and subsequently $1$ is not a good vertex.
• Let the starting vertex be $2$. We have two choices of traversing its neighbors. If we visit $3$ first, then the spanning tree will consist of edges ${2,3}$ and ${1,3}$, which is not equal to $T$. If we, however, visit $1$ first, then we can only continue to $3$ from here, and the spanning tree will consist of edges ${1, 2}$ and ${1,3}$, which equals to $T$. Hence, $2$ is a good vertex.
• The case when we start in the vertex $3$ is symmetrical to starting in $2$, and hence $3$ is a good vertex.

Therefore, the answer is $2$.

After the second update, the edge between $2$ and $3$ is removed, and $G = T$. It follows that the spanning tree generated by DFS will be always equal to $T$ independent of the choice of the starting vertex. Thus, the answer is $3$.

In the second sample, the set of good vertices after the corresponding query is:

• ${2, 3, 5}$
• ${3, 5}$
• ${3, 4, 5}$
• ${4, 5}$
• ${4, 5, 6}$
• ${5, 6}$

第一个样例如下图所示。

第一次更新后，图 $G$ 包含全部三条可能的边。一次 DFS 的结果如下：

• 设起始顶点为 $1$。对于顶点 $1$ 的邻接点，我们有两种排序方式：$[2, 3]$ 或 $[3, 2]$。
  • 若选择前者，则首先访问顶点 $2$；无论其邻接点如何排序，下一个被访问的顶点均为 $3$。因此，该 DFS 生成的生成树包含边 $\{1, 2\}$ 和 $\{2, 3\}$，不等于 $T$。
  • 若选择后者，则生成树包含边 $\{1, 3\}$ 和 $\{2, 3\}$。因此，不存在一种对各顶点邻接点的排序方式，使得 DFS 生成的生成树恰好为 $T$，故 $1$ 不是好顶点。
• 设起始顶点为 $2$。我们有两种遍历其邻接点的顺序。若先访问 $3$，则生成树包含边 $\{2, 3\}$ 和 $\{1, 3\}$，不等于 $T$；若先访问 $1$，则只能从 $1$ 继续访问 $3$，此时生成树包含边 $\{1, 2\}$ 和 $\{1, 3\}$，恰等于 $T$。因此，$2$ 是好顶点。
• 起始顶点为 $3$ 的情况与起始顶点为 $2$ 对称，故 $3$ 也是好顶点。

因此，答案为 $2$。

第二次更新后，边 $(2, 3)$ 被移除，此时 $G = T$。于是，无论选择哪个顶点作为起始点，DFS 生成的生成树恒为 $T$。因此，答案为 $3$。

第二个样例中，每次查询后的好顶点集合分别为：

• $\{2, 3, 5\}$
• $\{3, 5\}$
• $\{3, 4, 5\}$
• $\{4, 5\}$
• $\{4, 5, 6\}$
• $\{5, 6\}$