/*


*/
#include <set>
#include <cstdio>
#include <cctype>
#include <algorithm>
#define MAXIN 300000
#define gc() (SS==TT&&(TT=(SS=IN)+fread(IN,1,MAXIN,stdin),SS==TT)?EOF:*SS++)
typedef long long LL;
const int N=1e5+5;

int fa[N];
std::set<int> st[N];
std::pair<int,int> q[N];
char IN[MAXIN],*SS=IN,*TT=IN;

inline int read()
{
	int now=0;register char c=gc();
	for(;!isdigit(c);c=gc());
	for(;isdigit(c);now=now*10+c-'0',c=gc());
	return now;
}
int Find(int x)
{
	return x==fa[x]?x:fa[x]=Find(fa[x]);
}

int main()
{
	const int n=read();
	int h=0,t=0;
	for(int i=n-1<<1; i; --i)
	{
		int u=read(),v=read();
		if(u>v) std::swap(u,v);
		if(!st[u].count(v)) st[u].insert(v), st[v].insert(u);
		else q[t++]=std::make_pair(u,v);
	}
	for(int i=1; i<=n; ++i) fa[i]=i;
	while(h<t)
	{
		int x=Find(q[h].first),y=Find(q[h++].second);
		if(st[x].size()<st[y].size()) std::swap(x,y);//x<-y
		for(std::set<int>::iterator it=st[y].begin(); it!=st[y].end(); ++it)
			if(*it!=x)
			{
				int u=x,v=*it;
				st[v].erase(y);
				if(u>v) std::swap(u,v);
				if(!st[u].count(v)) st[u].insert(v), st[v].insert(u);
				else q[t++]=std::make_pair(u,v);		
			}
		st[x].erase(y), fa[y]=x;
	}
	puts(t==n-1?"YES":"NO");

	return 0;
}
### Prompt

Create a solution in cpp for 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 <set>
#include <cstdio>
#include <cctype>
#include <algorithm>
#define MAXIN 300000
#define gc() (SS==TT&&(TT=(SS=IN)+fread(IN,1,MAXIN,stdin),SS==TT)?EOF:*SS++)
typedef long long LL;
const int N=1e5+5;

int fa[N];
std::set<int> st[N];
std::pair<int,int> q[N];
char IN[MAXIN],*SS=IN,*TT=IN;

inline int read()
{
	int now=0;register char c=gc();
	for(;!isdigit(c);c=gc());
	for(;isdigit(c);now=now*10+c-'0',c=gc());
	return now;
}
int Find(int x)
{
	return x==fa[x]?x:fa[x]=Find(fa[x]);
}

int main()
{
	const int n=read();
	int h=0,t=0;
	for(int i=n-1<<1; i; --i)
	{
		int u=read(),v=read();
		if(u>v) std::swap(u,v);
		if(!st[u].count(v)) st[u].insert(v), st[v].insert(u);
		else q[t++]=std::make_pair(u,v);
	}
	for(int i=1; i<=n; ++i) fa[i]=i;
	while(h<t)
	{
		int x=Find(q[h].first),y=Find(q[h++].second);
		if(st[x].size()<st[y].size()) std::swap(x,y);//x<-y
		for(std::set<int>::iterator it=st[y].begin(); it!=st[y].end(); ++it)
			if(*it!=x)
			{
				int u=x,v=*it;
				st[v].erase(y);
				if(u>v) std::swap(u,v);
				if(!st[u].count(v)) st[u].insert(v), st[v].insert(u);
				else q[t++]=std::make_pair(u,v);		
			}
		st[x].erase(y), fa[y]=x;
	}
	puts(t==n-1?"YES":"NO");

	return 0;
}
```