題目連結：https://codeforces.com/contest/1037/problem/D

D. Valid BFS?

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The BFS algorithm is defined as follows.

1. Consider an undirected graph with vertices numbered from 11 to nn. Initialize qq as a new 
   queue containing only vertex 11, mark the vertex 11 as used.
2. Extract a vertex vv from the head of the queue qq.
3. Print the index of vertex vv.
4. Iterate in arbitrary order through all such vertices uu that uu is a neighbor of vv and is not marked yet as used. Mark the vertex uu as used and insert it into the tail of the queue qq.
5. If the queue is not empty, continue from step 2.
6. 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 11. The 

Input

The first line contains a single integer nn (1≤n≤2⋅1051≤n≤2⋅105) which denotes the number of nodes in the tree.

The following n−1n−1 lines describe the edges of the tree. Each of them contains two integers xx and yy (1≤x,y≤n1≤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 nn distinct integers a1,a2,…,ana1,a2,…,an (1≤ai≤n1≤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:

• 1,2,3,41,2,3,4,
• 1,3,2,41,3,2,4.

The ordering 1,2,4,31,2,4,3 doesn't correspond to any valid BFS order.

題目大意：給出一顆樹，節點1為根節點，再給出一段序列，如果這個序列能用 bfs 遍歷順序得出，輸出Yes 否則，輸出No;

要訓練將轉換問題的能力，如果按照所給的序列以個一個比對，可能很難實現，

程式碼程式碼程式碼程式碼程式碼程式碼程式碼程式碼程式碼程式碼程式碼程式碼程式碼

我給你講啊，這個人，大佬--->

#include<iostream>
#include<queue>
#include<vector>
#include<set>
using namespace std;
int  const  MAXN=2*1e5+10;
queue<int> q;
vector< set<int > > vec(MAXN);//定義一個vector陣列，裡面是set<int>型別;
int dis[MAXN];
int main(){
    int n;
    cin>>n;
    int a,b;
    for(int i=0;i<n-1;i++){
        cin>>a>>b;
        vec[a].insert(b);//無向圖存圖;
        vec[b].insert(a);
    }
    for(int i=0;i<n;i++){
        cin>>dis[i];
    }
    q.push(1);
    if(dis[0]!=1){//開頭一定是1，如果不是1，就沒有必要往下進行了；
        cout<<"No"<<endl;
        return 0;
    }
    int i=1;
    while(!q.empty()){
        int s=q.front();//取出佇列中的元素；
        q.pop();
        //vec[s].find(dis[i] 表示以點s為父節點遍歷所有他的孩子，再將每個孩子放入到佇列中；
        while(i<n && vec[s].find(dis[i])!=vec[s].end()){
            q.push(dis[i]);
            i++;//i一直往前走，把dis[i]的值放入佇列之中，
        }
    }
    if(i==n){//如果i能夠一直走到頭，說明給出的序列滿足bfs的定義方式；
        cout<<"Yes"<<endl;
    }else{
        cout<<"No"<<endl;
    }
    return 0;
}