#include<cstdio>
#include<map>
#include<set>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=100005;
int n;
int fa[MAXN];
int find(int x)
{
	if(fa[x]==x)
		return x;
	return fa[x]=find(fa[x]);
}
struct node
{
	int x,y;
	node(){}
	node(int xx,int yy)
	{
		x=xx,y=yy;
	}
};
queue<node>Q;
map<long long,int>a;
set<int>w[MAXN];
long long getid(int x,int y)
{
	if(x<y)
		swap(x,y);
	return 1ll*x*n+y;
}
void link(int x,int y)
{
	w[x].insert(y);
	w[y].insert(x);
	a[getid(x,y)]++;
	if(a[getid(x,y)]>=3)
	{
		printf("NO\n");
		exit(0);
	}
	if(a[getid(x,y)]==2)
		Q.push(node(x,y));
}
int main()
{
	//freopen("bluered.in","r",stdin);
	//freopen("bluered.out","w",stdout);
	scanf("%d",&n);
	for(int i=1;i<=2*n-2;i++)
	{
		int u,v;
		scanf("%d %d",&u,&v);
		link(u,v);
	}
	for(int i=1;i<=n;i++)
	{
		fa[i]=i;
	}
	for(int i=1;i<n;i++)
	{
		int x=0,y=0;
		while(1)
		{
			if(Q.empty())
			{
				printf("NO\n");
				return 0;
			}
			x=find(Q.front().x);
			y=find(Q.front().y);
			Q.pop();
			if(x!=y)
				break;
		}
		if(w[x].size()>w[y].size())
			swap(x,y);
		fa[x]=y;
		a.erase(getid(x,y));
		w[y].erase(x);
		for(set<int>::iterator it=w[x].begin();it!=w[x].end();it++)
		{
			int v=find(*it);
			if(v==y)
				continue;
			a.erase(getid(x,v));
			link(v,y);
			w[v].erase(x);
			w[x].erase(v);
		}
	}
	printf("YES\n");
	//fclose(stdin);
	//fclose(stdout);
}
### Prompt

Your challenge is to write 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<map>
#include<set>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=100005;
int n;
int fa[MAXN];
int find(int x)
{
	if(fa[x]==x)
		return x;
	return fa[x]=find(fa[x]);
}
struct node
{
	int x,y;
	node(){}
	node(int xx,int yy)
	{
		x=xx,y=yy;
	}
};
queue<node>Q;
map<long long,int>a;
set<int>w[MAXN];
long long getid(int x,int y)
{
	if(x<y)
		swap(x,y);
	return 1ll*x*n+y;
}
void link(int x,int y)
{
	w[x].insert(y);
	w[y].insert(x);
	a[getid(x,y)]++;
	if(a[getid(x,y)]>=3)
	{
		printf("NO\n");
		exit(0);
	}
	if(a[getid(x,y)]==2)
		Q.push(node(x,y));
}
int main()
{
	//freopen("bluered.in","r",stdin);
	//freopen("bluered.out","w",stdout);
	scanf("%d",&n);
	for(int i=1;i<=2*n-2;i++)
	{
		int u,v;
		scanf("%d %d",&u,&v);
		link(u,v);
	}
	for(int i=1;i<=n;i++)
	{
		fa[i]=i;
	}
	for(int i=1;i<n;i++)
	{
		int x=0,y=0;
		while(1)
		{
			if(Q.empty())
			{
				printf("NO\n");
				return 0;
			}
			x=find(Q.front().x);
			y=find(Q.front().y);
			Q.pop();
			if(x!=y)
				break;
		}
		if(w[x].size()>w[y].size())
			swap(x,y);
		fa[x]=y;
		a.erase(getid(x,y));
		w[y].erase(x);
		for(set<int>::iterator it=w[x].begin();it!=w[x].end();it++)
		{
			int v=find(*it);
			if(v==y)
				continue;
			a.erase(getid(x,v));
			link(v,y);
			w[v].erase(x);
			w[x].erase(v);
		}
	}
	printf("YES\n");
	//fclose(stdin);
	//fclose(stdout);
}
```