#include<bits/stdc++.h>
using namespace std;
#define pii pair<int,int>
#define mk make_pair
pii q[200001];
int h,t;
int n;
set<int> s[200001];
map<pii,int> mp;
int fa[200001];
int findfa(int x){
	return x==fa[x]?fa[x]:fa[x]=findfa(fa[x]);
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=(n-1)*2;++i){
		int u,v;
		scanf("%d%d",&u,&v);
		mp[mk(u,v)]++;mp[mk(v,u)]++;
		s[u].insert(v),s[v].insert(u);
		if(mp[mk(u,v)]==2)q[++t]=mk(u,v);
	}
	for(int i=1;i<=n;++i)fa[i]=i;
	set<int>::iterator id;
	while(h<t){
		int u=q[++h].first,v=q[h].second;
		u=findfa(u),v=findfa(v);
		s[u].erase(v),s[v].erase(u);
		if(s[u].size()<s[v].size())swap(u,v);
		for(id=s[v].begin();id!=s[v].end();++id){
			int to=(*id);
			mp[mk(v,to)]=mp[mk(to,v)]=0;
			s[to].erase(v);s[to].insert(u);
			mp[mk(u,to)]++,mp[mk(to,u)]++;
			if(mp[mk(u,to)]==2){
				q[++t]=mk(u,to);
			}
			s[u].insert(to);
		}
		fa[v]=u;
		s[v].clear();
	}
	int ans=0;
	for(int i=1;i<=n;++i)ans+=fa[i]==i;
	if(ans==1)puts("YES");
	else puts("NO");
}
### Prompt

In CPP, your task is to solve 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;
#define pii pair<int,int>
#define mk make_pair
pii q[200001];
int h,t;
int n;
set<int> s[200001];
map<pii,int> mp;
int fa[200001];
int findfa(int x){
	return x==fa[x]?fa[x]:fa[x]=findfa(fa[x]);
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=(n-1)*2;++i){
		int u,v;
		scanf("%d%d",&u,&v);
		mp[mk(u,v)]++;mp[mk(v,u)]++;
		s[u].insert(v),s[v].insert(u);
		if(mp[mk(u,v)]==2)q[++t]=mk(u,v);
	}
	for(int i=1;i<=n;++i)fa[i]=i;
	set<int>::iterator id;
	while(h<t){
		int u=q[++h].first,v=q[h].second;
		u=findfa(u),v=findfa(v);
		s[u].erase(v),s[v].erase(u);
		if(s[u].size()<s[v].size())swap(u,v);
		for(id=s[v].begin();id!=s[v].end();++id){
			int to=(*id);
			mp[mk(v,to)]=mp[mk(to,v)]=0;
			s[to].erase(v);s[to].insert(u);
			mp[mk(u,to)]++,mp[mk(to,u)]++;
			if(mp[mk(u,to)]==2){
				q[++t]=mk(u,to);
			}
			s[u].insert(to);
		}
		fa[v]=u;
		s[v].clear();
	}
	int ans=0;
	for(int i=1;i<=n;++i)ans+=fa[i]==i;
	if(ans==1)puts("YES");
	else puts("NO");
}
```