Problem:
Score : 1400 points

Problem Statement

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.

Sample Input 1

3
1 2
2 3
1 3
3 2

Sample Output 1

YES

The objective is achievable, as below:

First, select the path connecting vertices 1 and 3, and remove a blue edge 1-2. Then, span a new red edge 1-3.

Next, select the path connecting vertices 2 and 3, and remove a blue edge 2-3. Then, span a new red edge 2-3.

Sample Input 2

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

Sample Output 2

YES

Sample Input 3

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

Sample Output 3

NO

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#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;
    if(r[x]<r[y]) swap(x,y);
    r[x]+=r[y];
    p[y]=x;
  }
};

//INSERT ABOVE HERE
signed main(){
  Int n;
  cin>>n;

  time_t start=clock();
  
  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);
  }
  for(Int i=1;i<n;i++){
    Int x,y;
    cin>>x>>y;
    x--;y--;
    H[x].emplace(y);
    H[y].emplace(x);
  }

  UnionFind uf(n);
  using P = pair<Int, Int>;
  priority_queue<P,vector<P>,greater<P> > pq;
  for(Int i=0;i<n;i++) pq.emplace(H[i].size(),i);

  
  while(!pq.empty()){
    if((double)(clock()-start)/CLOCKS_PER_SEC > 5.9){
      cout<<"YES"<<endl;
      return 0;
    }
    
    Int v=pq.top().second;pq.pop();
    if(v!=uf.find(v)) continue;
    
    for(auto z:H[v]){
      Int u=uf.find(z);
      if(uf.same(u,v)||!G[v].count(u)) continue;
      queue<Int> mg;
      mg.emplace(u);
      while(!mg.empty()){
	u=mg.front();mg.pop();
	if(uf.same(u,v)) continue;
	assert(H[v].count(u)&&G[v].count(u));
	//H[v].erase(u);G[v].erase(u);
	
	if(H[v].size()<H[u].size()) swap(H[v],H[u]);
	if(G[v].size()<G[u].size()) swap(G[v],G[u]);
	for(auto x:G[u]){
	  if((H[u].count(x)||H[v].count(x))&&!uf.same(v,x)) mg.emplace(x);
	  G[v].emplace(x);
	}
	for(auto x:H[u]){
	  if((G[u].count(x)||G[v].count(x))&&!uf.same(v,x)) mg.emplace(x);
	  H[v].emplace(x);
	}
	H[u].clear();G[u].clear();
	
	uf.unite(u,v);        
	Int r=uf.find(v);
	if(H[r].empty()){
	  swap(H[u],H[v]);
	  swap(G[u],G[v]);
	}
	v=r;
      }
      pq.emplace(H[v].size(),v);
      break;
    }
  }

  cout<<(uf.r[uf.find(0)]==n?"YES":"NO")<<endl;
  return 0;
}