// luogu-judger-enable-o2
#include<cstdio>
#include<algorithm>
#include<queue>
#include<set>
using namespace std;
const int N=100011;
template<typename TP>inline void read(TP &kk)
{
	TP ret=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){ret=ret*10+ch-'0';ch=getchar();}
	kk=ret*f;
}
int n,fa[N];
int find(int x)
{
	if(fa[x]!=fa[fa[x]]) find(fa[x]);
	return fa[x]=fa[fa[x]];
}
struct shino{int x,y;shino(){}shino(int x,int y):x(x),y(y){}};
queue<shino> q;
set<int> to[N];
set<int>::iterator it,it2;
int counter;

int xi,yi,ti;
int main()
{
	read(n);for(int i=1;i<=n;i++) fa[i]=i;
	for(int i=3;i<=n*2;i++)
	{
		read(xi),read(yi);
		if(to[xi].find(yi)!=to[xi].end())
			q.push(shino(xi,yi));
		else to[xi].insert(yi),to[yi].insert(xi);
	}
	while(!q.empty())
	{
		shino g=q.front();q.pop();
		xi=g.x,yi=g.y;
		xi=find(xi),yi=find(yi);
		if(xi==yi) continue;
		to[xi].erase(yi),to[yi].erase(xi);
		counter++;
		if(to[xi].size()<to[yi].size()) swap(xi,yi);
		for(it=to[yi].begin();it!=to[yi].end();it++)
		{
			ti=*it;to[ti].erase(yi);
			if(to[xi].find(ti)!=to[xi].end())
				q.push(shino(xi,ti));
			else to[xi].insert(ti),to[ti].insert(xi);
		}
		fa[yi]=xi;
	}
	// counter=0;
	// for(int i=1;i<=n;i++) counter+=find(i)==i;
	if(counter==n-1) puts("YES");
	else puts("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
// luogu-judger-enable-o2
#include<cstdio>
#include<algorithm>
#include<queue>
#include<set>
using namespace std;
const int N=100011;
template<typename TP>inline void read(TP &kk)
{
	TP ret=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){ret=ret*10+ch-'0';ch=getchar();}
	kk=ret*f;
}
int n,fa[N];
int find(int x)
{
	if(fa[x]!=fa[fa[x]]) find(fa[x]);
	return fa[x]=fa[fa[x]];
}
struct shino{int x,y;shino(){}shino(int x,int y):x(x),y(y){}};
queue<shino> q;
set<int> to[N];
set<int>::iterator it,it2;
int counter;

int xi,yi,ti;
int main()
{
	read(n);for(int i=1;i<=n;i++) fa[i]=i;
	for(int i=3;i<=n*2;i++)
	{
		read(xi),read(yi);
		if(to[xi].find(yi)!=to[xi].end())
			q.push(shino(xi,yi));
		else to[xi].insert(yi),to[yi].insert(xi);
	}
	while(!q.empty())
	{
		shino g=q.front();q.pop();
		xi=g.x,yi=g.y;
		xi=find(xi),yi=find(yi);
		if(xi==yi) continue;
		to[xi].erase(yi),to[yi].erase(xi);
		counter++;
		if(to[xi].size()<to[yi].size()) swap(xi,yi);
		for(it=to[yi].begin();it!=to[yi].end();it++)
		{
			ti=*it;to[ti].erase(yi);
			if(to[xi].find(ti)!=to[xi].end())
				q.push(shino(xi,ti));
			else to[xi].insert(ti),to[ti].insert(xi);
		}
		fa[yi]=xi;
	}
	// counter=0;
	// for(int i=1;i<=n;i++) counter+=find(i)==i;
	if(counter==n-1) puts("YES");
	else puts("NO");
	return 0;
}
```