#include <iostream>
#include <cstdio>
#include <queue>
#include <map>
#include <set>
using namespace std;
inline int read(){int tot=1;char c=getchar();while(c!='-'&&(c<'0'||c>'9'))c=getchar();if(c=='-')tot=-1,c=getchar();
int sum=0;while(c<='9'&&c>='0')sum=sum*10+c-'0',c=getchar();return sum*tot;}
int fa[100005];
set<int> s[100005];
set<int>::iterator it;
map<pair<int,int>,int> mp;
queue<pair<int,int> > q;
int find(int x){
	if(fa[x]!=x)return fa[x]=find(fa[x]);
	return fa[x];
}
int main()
{
	int n=read();
	for(int i=1;i<=(n-1)*2;i++){
		int x=read(),y=read();
		s[x].insert(y);s[y].insert(x);
		mp[make_pair(x,y)]++;mp[make_pair(y,x)]++;
		if(mp[make_pair(x,y)]==2)q.push(make_pair(x,y));
	}
	for(int i=1;i<=n;i++)fa[i]=i;
	while(!q.empty()){
		int x=q.front().first,y=q.front().second;q.pop();
//		cout<<x<<" "<<y<<endl;
		x=find(x),y=find(y);
		s[x].erase(y);s[y].erase(x);
		if(s[x].size()<s[y].size())swap(x,y);
		for(it=s[y].begin();it!=s[y].end();it++){
//			cout<<"aa";
			int j=(*it);
			s[j].erase(y);//s[y].earse(j)这样子写GG
			mp[make_pair(y,j)]=0;mp[make_pair(j,y)]=0;
			s[j].insert(x);s[x].insert(j);
			mp[make_pair(x,j)]++;mp[make_pair(j,x)]++;
			if(mp[make_pair(j,x)]==2)q.push(make_pair(j,x));
		}
		fa[y]=x;
        s[y].clear();//
	}
	int sum=0;
	for(int i=1;i<=n;i++){
		if(fa[i]==i)sum++;
//		cout<<fa[i]<<" ";
	}
	if(sum==1)puts("YES");
	else puts("NO");
	return 0;
}
//反过来做的思路 原图删除的最后一条边在新图中也以同样形式出现，维护一下就好了
### 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 <iostream>
#include <cstdio>
#include <queue>
#include <map>
#include <set>
using namespace std;
inline int read(){int tot=1;char c=getchar();while(c!='-'&&(c<'0'||c>'9'))c=getchar();if(c=='-')tot=-1,c=getchar();
int sum=0;while(c<='9'&&c>='0')sum=sum*10+c-'0',c=getchar();return sum*tot;}
int fa[100005];
set<int> s[100005];
set<int>::iterator it;
map<pair<int,int>,int> mp;
queue<pair<int,int> > q;
int find(int x){
	if(fa[x]!=x)return fa[x]=find(fa[x]);
	return fa[x];
}
int main()
{
	int n=read();
	for(int i=1;i<=(n-1)*2;i++){
		int x=read(),y=read();
		s[x].insert(y);s[y].insert(x);
		mp[make_pair(x,y)]++;mp[make_pair(y,x)]++;
		if(mp[make_pair(x,y)]==2)q.push(make_pair(x,y));
	}
	for(int i=1;i<=n;i++)fa[i]=i;
	while(!q.empty()){
		int x=q.front().first,y=q.front().second;q.pop();
//		cout<<x<<" "<<y<<endl;
		x=find(x),y=find(y);
		s[x].erase(y);s[y].erase(x);
		if(s[x].size()<s[y].size())swap(x,y);
		for(it=s[y].begin();it!=s[y].end();it++){
//			cout<<"aa";
			int j=(*it);
			s[j].erase(y);//s[y].earse(j)这样子写GG
			mp[make_pair(y,j)]=0;mp[make_pair(j,y)]=0;
			s[j].insert(x);s[x].insert(j);
			mp[make_pair(x,j)]++;mp[make_pair(j,x)]++;
			if(mp[make_pair(j,x)]==2)q.push(make_pair(j,x));
		}
		fa[y]=x;
        s[y].clear();//
	}
	int sum=0;
	for(int i=1;i<=n;i++){
		if(fa[i]==i)sum++;
//		cout<<fa[i]<<" ";
	}
	if(sum==1)puts("YES");
	else puts("NO");
	return 0;
}
//反过来做的思路 原图删除的最后一条边在新图中也以同样形式出现，维护一下就好了
```