#include<cstdio>
#include<map>
#include<set>
#include<queue>
#include<algorithm>
#define MAXN 100006
using namespace std;
int n,fa[MAXN],u,v,top,p[MAXN];
map<int,int> h[MAXN];
set<int> e[MAXN];
queue<pair<int,int> > q;
void link(int u,int v)
{
	e[u].insert(v);
	e[v].insert(u);
	if(u>v) swap(u,v);
	++h[u][v];
	if(h[u][v]==2) q.push(make_pair(u,v));
	if(h[u][v]>2)
	{
		printf("NO");
		exit(0);
	}
}
void cut(int u,int v)
{
	e[u].erase(v);
	if(u>v) swap(u,v);
	h[u].erase(v);
}
int find(int x)
{
	if(x!=fa[x]) fa[x]=find(fa[x]);
	return fa[x];
}
int main()
{
	//freopen("bluered.in","r",stdin);
	//freopen("bluered.out","w",stdout);
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		fa[i]=i;
	for(int i=1;i<n;i++)
	{
		scanf("%d%d",&u,&v);
		link(u,v);
	}
	for(int i=1;i<n;i++)
	{
		scanf("%d%d",&u,&v);
		link(u,v);
	}
	for(int i=1;i<n;i++)
	{
		do
		{
			if(q.empty())
			{
				printf("NO");
				return 0;
			}
			u=q.front().first;
			v=q.front().second;
			u=find(u);
			v=find(v);
			q.pop();
		}while(u==v);
		if(e[u].size()>e[v].size()) swap(u,v);
		fa[u]=v;
		cut(v,u);
		top=0;		
		for(set<int>::iterator it=e[u].begin();it!=e[u].end();it++)
			p[++top]=*it;
		for(int i=1;i<=top;i++)
		{
			int x=p[i];
			x=find(x);
			if(x==v) continue;
			link(x,v);
			cut(x,u);
		}
		e[u].clear();
	}
	printf("YES");
}
### Prompt

Please create a solution in cpp 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<cstdio>
#include<map>
#include<set>
#include<queue>
#include<algorithm>
#define MAXN 100006
using namespace std;
int n,fa[MAXN],u,v,top,p[MAXN];
map<int,int> h[MAXN];
set<int> e[MAXN];
queue<pair<int,int> > q;
void link(int u,int v)
{
	e[u].insert(v);
	e[v].insert(u);
	if(u>v) swap(u,v);
	++h[u][v];
	if(h[u][v]==2) q.push(make_pair(u,v));
	if(h[u][v]>2)
	{
		printf("NO");
		exit(0);
	}
}
void cut(int u,int v)
{
	e[u].erase(v);
	if(u>v) swap(u,v);
	h[u].erase(v);
}
int find(int x)
{
	if(x!=fa[x]) fa[x]=find(fa[x]);
	return fa[x];
}
int main()
{
	//freopen("bluered.in","r",stdin);
	//freopen("bluered.out","w",stdout);
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		fa[i]=i;
	for(int i=1;i<n;i++)
	{
		scanf("%d%d",&u,&v);
		link(u,v);
	}
	for(int i=1;i<n;i++)
	{
		scanf("%d%d",&u,&v);
		link(u,v);
	}
	for(int i=1;i<n;i++)
	{
		do
		{
			if(q.empty())
			{
				printf("NO");
				return 0;
			}
			u=q.front().first;
			v=q.front().second;
			u=find(u);
			v=find(v);
			q.pop();
		}while(u==v);
		if(e[u].size()>e[v].size()) swap(u,v);
		fa[u]=v;
		cut(v,u);
		top=0;		
		for(set<int>::iterator it=e[u].begin();it!=e[u].end();it++)
			p[++top]=*it;
		for(int i=1;i<=top;i++)
		{
			int x=p[i];
			x=find(x);
			if(x==v) continue;
			link(x,v);
			cut(x,u);
		}
		e[u].clear();
	}
	printf("YES");
}
```