The BFS algorithm is defined as follows.

1. Consider an undirected graph with vertices numbered from 1

to n. Initialize q as a new queue containing only vertex 1, mark the vertex 1

• as used.
• Extract a vertex v
• from the head of the queue q
• .
• Print the index of vertex v
• .
• Iterate in arbitrary order through all such vertices u
• that u is a neighbor of v and is not marked yet as used. Mark the vertex u as used and insert it into the tail of the queue q
  1. .
  2. If the queue is not empty, continue from step 2.
  3. Otherwise finish.

  Since the order of choosing neighbors of each vertex can vary, it turns out that there may be multiple sequences which BFS can print.

  In this problem you need to check whether a given sequence corresponds to some valid BFS traversal of the given tree starting from vertex 1

  . The tree is an undirected graph, such that there is exactly one simple path between any two vertices.

  Input

  The first line contains a single integer n

  (1≤n≤2⋅105

  ) which denotes the number of nodes in the tree.

  The following n−1

  lines describe the edges of the tree. Each of them contains two integers x and y (1≤x,y≤n

  ) — the endpoints of the corresponding edge of the tree. It is guaranteed that the given graph is a tree.

  The last line contains n

  distinct integers a1,a2,…,an (1≤ai≤n

  ) — the sequence to check.

  Output

  Print "Yes" (quotes for clarity) if the sequence corresponds to some valid BFS traversal of the given tree and "No" (quotes for clarity) otherwise.

  You can print each letter in any case (upper or lower).

  Examples

  Input

  Copy

  4
  1 2
  1 3
  2 4
  1 2 3 4
  

  Output

  Copy

  Yes

  Input

  Copy

  4
  1 2
  1 3
  2 4
  1 2 4 3
  

  Output

  Copy

  No

  Note

  Both sample tests have the same tree in them.

  In this tree, there are two valid BFS orderings:

  • 1,2,3,4
• ,
• 1,3,2,4

• .

The ordering 1,2,4,3

doesn't correspond to any valid BFS order.

題意：給了一顆n節點，n-1條邊的樹，進行BFS遍歷，問遍歷順序是否可能是所給陣列的順序

思路：先對構樹的鄰接表通過序列中的數的次序進行排序，再直接對樹bfs，看其結果是否相同即可

#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
const int N = 2e5+5;
vector<int>G[N];
int arr[N],id[N];
bool vis[N];
void init(){
	memset(vis,0,sizeof(vis));
	for(int i=0;i<N;i++)
		G[i].clear();
}
bool cmp(int a,int b){
	return id[a]<id[b];
}
int BFS(){
	vis[1]-=1;
	int cnt=1;
	queue<int>p;
	p.push(1);
	while(!p.empty()){
		int u=p.front();
		p.pop();
		if(arr[cnt]!=u) return 0;
		else cnt++;
		for(int i=0;i<G[u].size();i++){
			if(!vis[G[u][i]]){
				vis[G[u][i]]=1;
				p.push(G[u][i]);
			}
		}
	}
	return 1;
}
int main(){
	int n;
	while(~scanf("%d",&n)){
		init();
		for(int i=1;i<n;i++){
			int u,v;
			scanf("%d%d",&u,&v);
			G[u].push_back(v);
			G[v].push_back(u);
		}
		for(int i=1;i<=n;i++){
			scanf("%d",&arr[i]);
			id[arr[i]]=i;
		}
		for(int i=1;i<=n;i++)
			sort(G[i].begin(),G[i].end(),cmp);
		if(BFS()) printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}