#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<vector>
#include<map>
#include<set>
using namespace std;
set<int>s[100001];
map<pair<int,int>,int>m;
queue<pair<int,int> >q;
#define mp make_pair
#define IT set<int>::iterator
int fa[100001];
int find(int x){return (fa[x]==x)?fa[x]:find(fa[x]);}
signed main()
{
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		fa[i]=i;
	for(int i=1;i<=2*n-2;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		s[u].insert(v);
		s[v].insert(u);
		m[mp(u,v)]++,m[mp(v,u)]++;
		if(m[mp(u,v)]==2)
			q.push(mp(u,v));
	}
	while(!q.empty())
	{
		int u=q.front().first;
		int v=q.front().second;
		q.pop();
		u=find(u),v=find(v);
		if(u!=v)
		{
			if(s[v].size()>s[u].size())swap(u,v);
			s[u].erase(v);
			s[v].erase(u);
			for(IT it=s[v].begin();it!=s[v].end();++it)
			{
				int to=(*it);
				m[mp(to,v)]=m[mp(v,to)]=0;
				s[to].erase(v);
				s[to].insert(u);
				s[u].insert(to);
				m[mp(u,to)]++,m[mp(to,u)]++;
				if(m[mp(u,to)]==2)
					q.push(mp(u,to));
				if(m[mp(to,u)]==2)
					q.push(mp(to,u));
			}
			fa[v]=u;
			s[v].clear();
		}
	}
	for(int i=2;i<=n;i++)
		if(find(i)!=find(i-1))
		{
			printf("NO");
			return 0;
		}
		printf("YES");
    return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<vector>
#include<map>
#include<set>
using namespace std;
set<int>s[100001];
map<pair<int,int>,int>m;
queue<pair<int,int> >q;
#define mp make_pair
#define IT set<int>::iterator
int fa[100001];
int find(int x){return (fa[x]==x)?fa[x]:find(fa[x]);}
signed main()
{
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		fa[i]=i;
	for(int i=1;i<=2*n-2;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		s[u].insert(v);
		s[v].insert(u);
		m[mp(u,v)]++,m[mp(v,u)]++;
		if(m[mp(u,v)]==2)
			q.push(mp(u,v));
	}
	while(!q.empty())
	{
		int u=q.front().first;
		int v=q.front().second;
		q.pop();
		u=find(u),v=find(v);
		if(u!=v)
		{
			if(s[v].size()>s[u].size())swap(u,v);
			s[u].erase(v);
			s[v].erase(u);
			for(IT it=s[v].begin();it!=s[v].end();++it)
			{
				int to=(*it);
				m[mp(to,v)]=m[mp(v,to)]=0;
				s[to].erase(v);
				s[to].insert(u);
				s[u].insert(to);
				m[mp(u,to)]++,m[mp(to,u)]++;
				if(m[mp(u,to)]==2)
					q.push(mp(u,to));
				if(m[mp(to,u)]==2)
					q.push(mp(to,u));
			}
			fa[v]=u;
			s[v].clear();
		}
	}
	for(int i=2;i<=n;i++)
		if(find(i)!=find(i-1))
		{
			printf("NO");
			return 0;
		}
		printf("YES");
    return 0;
}

```