#include<cstdio>
#include<map>
#include<queue>
#include<set>
#include<algorithm>
using namespace std;
#define PII pair<int,int>
#define MAXN 100010
map<PII,int>G;
queue<PII>Q;
set<int>A[MAXN];
set<int>::iterator it;
int n,fa[MAXN];
int Find(int x)
{
	return fa[x]==x?x:fa[x]=Find(fa[x]);
}
int main()
{
	scanf("%d",&n);
	int u,v;
	for(int i=1;i<=2*n-2;i++)
	{
		scanf("%d%d",&u,&v);
		A[u].insert(v),A[v].insert(u);
		G[PII(u,v)]++;G[PII(v,u)]++;
		if(G[PII(u,v)]==2)
			Q.push(PII(u,v));
	}
	for(int i=1;i<=n;i++)
		fa[i]=i;
	while(!Q.empty())
	{
		int u=Q.front().first,v=Q.front().second;Q.pop();
		u=Find(u),v=Find(v);
		A[u].erase(v),A[v].erase(u);
		if(A[u].size()>A[v].size()) swap(u,v);
		for(it=A[u].begin();it!=A[u].end();it++)
		{
			int x=(*it);
			G[PII(x,u)]=G[PII(u,x)]=0;
			G[PII(x,v)]++,G[PII(v,x)]++;
			if(G[PII(x,v)]==2)
				Q.push(PII(x,v));
			A[x].erase(u);A[x].insert(v);
			A[v].insert(x);
		}
		fa[u]=v;
		A[u].clear();
	}
	int cnt=0;
	for(int i=1;i<=n;i++)
		cnt+=(fa[i]==i);
	if(cnt==1) printf("YES\n");
	else printf("NO\n");
}
### 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<queue>
#include<set>
#include<algorithm>
using namespace std;
#define PII pair<int,int>
#define MAXN 100010
map<PII,int>G;
queue<PII>Q;
set<int>A[MAXN];
set<int>::iterator it;
int n,fa[MAXN];
int Find(int x)
{
	return fa[x]==x?x:fa[x]=Find(fa[x]);
}
int main()
{
	scanf("%d",&n);
	int u,v;
	for(int i=1;i<=2*n-2;i++)
	{
		scanf("%d%d",&u,&v);
		A[u].insert(v),A[v].insert(u);
		G[PII(u,v)]++;G[PII(v,u)]++;
		if(G[PII(u,v)]==2)
			Q.push(PII(u,v));
	}
	for(int i=1;i<=n;i++)
		fa[i]=i;
	while(!Q.empty())
	{
		int u=Q.front().first,v=Q.front().second;Q.pop();
		u=Find(u),v=Find(v);
		A[u].erase(v),A[v].erase(u);
		if(A[u].size()>A[v].size()) swap(u,v);
		for(it=A[u].begin();it!=A[u].end();it++)
		{
			int x=(*it);
			G[PII(x,u)]=G[PII(u,x)]=0;
			G[PII(x,v)]++,G[PII(v,x)]++;
			if(G[PII(x,v)]==2)
				Q.push(PII(x,v));
			A[x].erase(u);A[x].insert(v);
			A[v].insert(x);
		}
		fa[u]=v;
		A[u].clear();
	}
	int cnt=0;
	for(int i=1;i<=n;i++)
		cnt+=(fa[i]==i);
	if(cnt==1) printf("YES\n");
	else printf("NO\n");
}
```