#include<bits/stdc++.h>
using namespace std;
#define maxn 100010

#define fi first
#define se second
typedef pair <int ,int > par;

template < typename tp > void read ( tp & dig ){
	char ch=getchar();dig=0;int flag=0;
	while(!isdigit(ch)){ if(ch=='-')flag=1;ch=getchar(); }
	while(isdigit(ch))dig=dig*10+ch-'0',ch=getchar();
	if(flag)dig=-dig;
}

set <int > ::iterator it;
set <int > s[maxn];
map <par ,int > mp;
queue <par > que;
int n,fa[maxn];

inline int find(int x){
	return x==fa[x]?x:fa[x]=find(fa[x]);
}

int main()
{

	read(n);int nn=2*n-2;
	for(int i=1;i<=n;++i)fa[i]=i;
	for(int i=1,x,y;i<=nn;++i){
		read(x);read(y);
		s[x].insert(y),s[y].insert(x);
		++mp[par(x,y)],++mp[par(y,x)];
		if(mp[par(x,y)]==2)que.push(par(x,y));
	}while(!que.empty()){
		par tmp=que.front();que.pop();
		int x=find(tmp.fi),y=find(tmp.se);
		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++){
			int to=*(it);mp[par(to,y)]=mp[par(y,to)]=0;
			s[to].erase(y),s[to].insert(x);
			++mp[par(to,x)],++mp[par(x,to)];
			if(mp[par(to,x)]==2)que.push(par(to,x));
			s[x].insert(to);
		}fa[y]=x;s[y].clear();
	}int sum=0;
	for(int i=1;i<=n;++i)
		sum+=(find(i)==i);
	if(sum==1)printf("YES\n");
	else printf("NO\n");
	return 0;
	
}

### 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<bits/stdc++.h>
using namespace std;
#define maxn 100010

#define fi first
#define se second
typedef pair <int ,int > par;

template < typename tp > void read ( tp & dig ){
	char ch=getchar();dig=0;int flag=0;
	while(!isdigit(ch)){ if(ch=='-')flag=1;ch=getchar(); }
	while(isdigit(ch))dig=dig*10+ch-'0',ch=getchar();
	if(flag)dig=-dig;
}

set <int > ::iterator it;
set <int > s[maxn];
map <par ,int > mp;
queue <par > que;
int n,fa[maxn];

inline int find(int x){
	return x==fa[x]?x:fa[x]=find(fa[x]);
}

int main()
{

	read(n);int nn=2*n-2;
	for(int i=1;i<=n;++i)fa[i]=i;
	for(int i=1,x,y;i<=nn;++i){
		read(x);read(y);
		s[x].insert(y),s[y].insert(x);
		++mp[par(x,y)],++mp[par(y,x)];
		if(mp[par(x,y)]==2)que.push(par(x,y));
	}while(!que.empty()){
		par tmp=que.front();que.pop();
		int x=find(tmp.fi),y=find(tmp.se);
		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++){
			int to=*(it);mp[par(to,y)]=mp[par(y,to)]=0;
			s[to].erase(y),s[to].insert(x);
			++mp[par(to,x)],++mp[par(x,to)];
			if(mp[par(to,x)]==2)que.push(par(to,x));
			s[x].insert(to);
		}fa[y]=x;s[y].clear();
	}int sum=0;
	for(int i=1;i<=n;++i)
		sum+=(find(i)==i);
	if(sum==1)printf("YES\n");
	else printf("NO\n");
	return 0;
	
}

```