#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<ctime>
#include<utility>
#include<map>
#include<queue>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
int n;
map<ll,int> a;
queue<pii> q;
set<int> l[100010];
int f[100010];
int find(int x)
{
	return f[x]==x?x:f[x]=find(f[x]);
}
ll getid(int x,int y)
{
	if(x>y)
		swap(x,y);
	return ll(x)*n+y;
}
void link(int x,int y)
{
	l[x].insert(y);
	l[y].insert(x);
	ll v=getid(x,y);
	a[v]++;
	if(a[v]==3)
	{
		printf("NO\n");
		exit(0);
	}
	if(a[v]==2)
		q.push(pii(x,y));
}
int main()
{
	scanf("%d",&n);
	int i,x,y;
	for(i=1;i<=2*n-2;i++)
	{
		scanf("%d%d",&x,&y);
		link(x,y);
	}
	for(i=1;i<=n;i++)
		f[i]=i;
	for(i=1;i<n;i++)
	{
		while(1)
		{
			if(q.empty())
			{
				printf("NO\n");
				return 0;
			}
			x=q.front().first;
			y=q.front().second;
			x=find(x);
			y=find(y);
			q.pop();
			if(x!=y)
				break;
		}
		if(l[x].size()>l[y].size())
			swap(x,y);
		f[x]=y;
		a.erase(getid(x,y));
		l[y].erase(x);
		for(auto v:l[x])
		{
			v=find(v);
			if(v==y)
				continue;
			link(v,y);
			l[v].erase(x);
			a.erase(getid(x,v));
		}
		l[x].clear();
	}
	printf("YES\n");
	return 0;
}
### Prompt

Generate a CPP solution 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<cstring>
#include<algorithm>
#include<cstdlib>
#include<ctime>
#include<utility>
#include<map>
#include<queue>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
int n;
map<ll,int> a;
queue<pii> q;
set<int> l[100010];
int f[100010];
int find(int x)
{
	return f[x]==x?x:f[x]=find(f[x]);
}
ll getid(int x,int y)
{
	if(x>y)
		swap(x,y);
	return ll(x)*n+y;
}
void link(int x,int y)
{
	l[x].insert(y);
	l[y].insert(x);
	ll v=getid(x,y);
	a[v]++;
	if(a[v]==3)
	{
		printf("NO\n");
		exit(0);
	}
	if(a[v]==2)
		q.push(pii(x,y));
}
int main()
{
	scanf("%d",&n);
	int i,x,y;
	for(i=1;i<=2*n-2;i++)
	{
		scanf("%d%d",&x,&y);
		link(x,y);
	}
	for(i=1;i<=n;i++)
		f[i]=i;
	for(i=1;i<n;i++)
	{
		while(1)
		{
			if(q.empty())
			{
				printf("NO\n");
				return 0;
			}
			x=q.front().first;
			y=q.front().second;
			x=find(x);
			y=find(y);
			q.pop();
			if(x!=y)
				break;
		}
		if(l[x].size()>l[y].size())
			swap(x,y);
		f[x]=y;
		a.erase(getid(x,y));
		l[y].erase(x);
		for(auto v:l[x])
		{
			v=find(v);
			if(v==y)
				continue;
			link(v,y);
			l[v].erase(x);
			a.erase(getid(x,v));
		}
		l[x].clear();
	}
	printf("YES\n");
	return 0;
}
```