#include<bits/stdc++.h>
#define pb push_back
#define fi first
#define se second

using namespace std;
typedef pair<int,int> pi;

const int N=100005;

int n,m,fa[N];
set<int> e[N];
set<pi> o;
inline void add(int u,int v){
	auto it=e[u].find(v);
	if(it!=e[u].end()){
		e[u].erase(it);
		if(u>v)swap(u,v);
		o.insert({u,v});
	}else e[u].insert(v);
}
int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}

int main(){
	scanf("%d",&n);
	for(int i=2,u,v;i<2*n;i++)scanf("%d%d",&u,&v),add(u,v),add(v,u);
	for(int i=1;i<=n;i++)fa[i]=i;
	while(o.size()){
		auto p=o.begin();
		int u=find(p->fi),v=find(p->se);
		o.erase(p);
		if(e[u].size()>e[v].size())swap(u,v);fa[u]=v;
		for(int k:e[u])e[k].erase(u),add(v,k),add(k,v);
		e[u].clear();
		m++;
	}
	if(m==n-1)puts("YES");else puts("NO");
	return 0;
}
### Prompt

Generate 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>
#define pb push_back
#define fi first
#define se second

using namespace std;
typedef pair<int,int> pi;

const int N=100005;

int n,m,fa[N];
set<int> e[N];
set<pi> o;
inline void add(int u,int v){
	auto it=e[u].find(v);
	if(it!=e[u].end()){
		e[u].erase(it);
		if(u>v)swap(u,v);
		o.insert({u,v});
	}else e[u].insert(v);
}
int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}

int main(){
	scanf("%d",&n);
	for(int i=2,u,v;i<2*n;i++)scanf("%d%d",&u,&v),add(u,v),add(v,u);
	for(int i=1;i<=n;i++)fa[i]=i;
	while(o.size()){
		auto p=o.begin();
		int u=find(p->fi),v=find(p->se);
		o.erase(p);
		if(e[u].size()>e[v].size())swap(u,v);fa[u]=v;
		for(int k:e[u])e[k].erase(u),add(v,k),add(k,v);
		e[u].clear();
		m++;
	}
	if(m==n-1)puts("YES");else puts("NO");
	return 0;
}
```