#include<bits/stdc++.h>
#define N 100010
using namespace std;
struct data
{
	int x,y;
};
set<data> s;
set<int> ss[N];
int n,fa[N],x,y;
bool operator <(data a,data b)
{
	if(a.x!=b.x)
	{
		return a.x<b.x;
	}
	return a.y<b.y;
}
int findfa(int x)
{
	if(fa[x]==x)
	{
		return x;
	}
	return fa[x]=findfa(fa[x]);
}
void adde(int x,int y)
{
	set<int>::iterator it=ss[x].find(y);
	if(it==ss[x].end())
	{
		ss[x].insert(y);
	}else{
		ss[x].erase(y);
		if(x>y)
		{
			swap(x,y);
		}
		s.insert((data){
			x,y
		});
	}
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		scanf("%d%d",&x,&y);
		adde(x,y);
		adde(y,x);
	}
	for(int i=1;i<n;i++)
	{
		scanf("%d%d",&x,&y);
		adde(x,y);
		adde(y,x);
	}
	for(int i=1;i<=n;i++)
	{
		fa[i]=i;
	}
	while(!s.empty())
	{
		set<data>::iterator it=s.begin();
		set<int>::iterator its;
		int x=findfa((*it).x),y=findfa((*it).y);
		s.erase(it);
		if(ss[x].size()>ss[y].size())
		{
			swap(x,y);
		}
		fa[x]=y;
		for(its=ss[x].begin();its!=ss[x].end();its++)
		{
			ss[*its].erase(x);
			adde(*its,y);
			adde(y,*its);
		}
		ss[x].clear();
		n--;
	}
	if(n==1)
	{
		puts("YES");
	}else{
		puts("NO");
	}
	return 0;
}
### Prompt

Develop a solution in cpp to the problem described below:

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>
#define N 100010
using namespace std;
struct data
{
	int x,y;
};
set<data> s;
set<int> ss[N];
int n,fa[N],x,y;
bool operator <(data a,data b)
{
	if(a.x!=b.x)
	{
		return a.x<b.x;
	}
	return a.y<b.y;
}
int findfa(int x)
{
	if(fa[x]==x)
	{
		return x;
	}
	return fa[x]=findfa(fa[x]);
}
void adde(int x,int y)
{
	set<int>::iterator it=ss[x].find(y);
	if(it==ss[x].end())
	{
		ss[x].insert(y);
	}else{
		ss[x].erase(y);
		if(x>y)
		{
			swap(x,y);
		}
		s.insert((data){
			x,y
		});
	}
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		scanf("%d%d",&x,&y);
		adde(x,y);
		adde(y,x);
	}
	for(int i=1;i<n;i++)
	{
		scanf("%d%d",&x,&y);
		adde(x,y);
		adde(y,x);
	}
	for(int i=1;i<=n;i++)
	{
		fa[i]=i;
	}
	while(!s.empty())
	{
		set<data>::iterator it=s.begin();
		set<int>::iterator its;
		int x=findfa((*it).x),y=findfa((*it).y);
		s.erase(it);
		if(ss[x].size()>ss[y].size())
		{
			swap(x,y);
		}
		fa[x]=y;
		for(its=ss[x].begin();its!=ss[x].end();its++)
		{
			ss[*its].erase(x);
			adde(*its,y);
			adde(y,*its);
		}
		ss[x].clear();
		n--;
	}
	if(n==1)
	{
		puts("YES");
	}else{
		puts("NO");
	}
	return 0;
}
```