#include<bits/stdc++.h>
using namespace std;
using Int = long long;

struct UnionFind{
  Int n;
  vector<Int> r,p;
  UnionFind(){}
  UnionFind(Int sz):n(sz),r(sz,1),p(sz,0){iota(p.begin(),p.end(),0);}
  Int find(Int x){
    return (x==p[x]?x:p[x]=find(p[x]));
  }
  bool same(Int x,Int y){
    return find(x)==find(y);
  }
  void unite(Int x,Int y){
    x=find(x);y=find(y);
    if(x==y) return;
    r[x]+=r[y];
    p[y]=x;
  }
};

//INSERT ABOVE HERE
signed main(){
  Int n;
  cin>>n;
  
  vector<set<Int> > G(n),H(n);
  for(Int i=1;i<n;i++){
    Int x,y;
    cin>>x>>y;
    x--;y--;
    G[x].emplace(y);
    G[y].emplace(x);
  }
  using P = pair<Int, Int>;
  queue<P> q;
  for(Int i=1;i<n;i++){
    Int x,y;
    cin>>x>>y;
    x--;y--;
    if(G[x].count(y)) q.emplace(x,y);
    H[x].emplace(y);
    H[y].emplace(x);
  }

  UnionFind uf(n);
  while(!q.empty()){
    Int x,y;
    tie(x,y)=q.front();q.pop();
    if(uf.same(x,y)) continue;
    x=uf.find(x);y=uf.find(y);
    
    uf.unite(x,y);
    
    for(auto z:H[y]){
      if(G[x].count(z)) q.emplace(x,z);
      H[x].emplace(z);
      H[z].emplace(x);
    }    
    for(auto z:G[y]){
      if(H[x].count(z)) q.emplace(x,z);
      G[x].emplace(z);
      G[z].emplace(x);
    }
  }
  cout<<(uf.r[uf.find(0)]==n?"YES":"NO")<<endl;
  return 0;
}

### Prompt

Please formulate a cpp solution to the following problem:

There is a tree with N vertices numbered 1 through N. The i-th of the N-1 edges connects vertices a_i and b_i.

Initially, each edge is painted blue. Takahashi will convert this blue tree into a red tree, by performing the following operation N-1 times:

* Select a simple path that consists of only blue edges, and remove one of those edges.
* Then, span a new red edge between the two endpoints of the selected path.



His objective is to obtain a tree that has a red edge connecting vertices c_i and d_i, for each i.

Determine whether this is achievable.

Constraints

* 2 ≤ N ≤ 10^5
* 1 ≤ a_i,b_i,c_i,d_i ≤ N
* a_i ≠ b_i
* c_i ≠ d_i
* Both input graphs are trees.

Input

Input is given from Standard Input in the following format:


N
a_1 b_1
:
a_{N-1} b_{N-1}
c_1 d_1
:
c_{N-1} d_{N-1}


Output

Print `YES` if the objective is achievable; print `NO` otherwise.

Examples

Input

3
1 2
2 3
1 3
3 2


Output

YES


Input

5
1 2
2 3
3 4
4 5
3 4
2 4
1 4
1 5


Output

YES


Input

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


Output

NO

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
using Int = long long;

struct UnionFind{
  Int n;
  vector<Int> r,p;
  UnionFind(){}
  UnionFind(Int sz):n(sz),r(sz,1),p(sz,0){iota(p.begin(),p.end(),0);}
  Int find(Int x){
    return (x==p[x]?x:p[x]=find(p[x]));
  }
  bool same(Int x,Int y){
    return find(x)==find(y);
  }
  void unite(Int x,Int y){
    x=find(x);y=find(y);
    if(x==y) return;
    r[x]+=r[y];
    p[y]=x;
  }
};

//INSERT ABOVE HERE
signed main(){
  Int n;
  cin>>n;
  
  vector<set<Int> > G(n),H(n);
  for(Int i=1;i<n;i++){
    Int x,y;
    cin>>x>>y;
    x--;y--;
    G[x].emplace(y);
    G[y].emplace(x);
  }
  using P = pair<Int, Int>;
  queue<P> q;
  for(Int i=1;i<n;i++){
    Int x,y;
    cin>>x>>y;
    x--;y--;
    if(G[x].count(y)) q.emplace(x,y);
    H[x].emplace(y);
    H[y].emplace(x);
  }

  UnionFind uf(n);
  while(!q.empty()){
    Int x,y;
    tie(x,y)=q.front();q.pop();
    if(uf.same(x,y)) continue;
    x=uf.find(x);y=uf.find(y);
    
    uf.unite(x,y);
    
    for(auto z:H[y]){
      if(G[x].count(z)) q.emplace(x,z);
      H[x].emplace(z);
      H[z].emplace(x);
    }    
    for(auto z:G[y]){
      if(H[x].count(z)) q.emplace(x,z);
      G[x].emplace(z);
      G[z].emplace(x);
    }
  }
  cout<<(uf.r[uf.find(0)]==n?"YES":"NO")<<endl;
  return 0;
}

```