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

Solution (in CPP):

Solution:
#include<iostream>
#include<map>
#include<set>
#include<queue>
#include<algorithm>
using namespace std;
const int N=100005;
int f[N];
int getf(int x)
{
	return f[x]==x?x:f[x]=getf(f[x]);
}
set<int> mat[N];
int main()
{
	ios::sync_with_stdio(false);
	int n;
	cin>>n;
	map<pair<int,int>,int> occur;
	queue<pair<int,int> > Q;
	for(int i=1;i<=n*2-2;i++)
	{
		int u,v;
		cin>>u>>v;
		mat[u].insert(v);
		mat[v].insert(u);
		if(u>v)
			swap(u,v);
		if(++occur[make_pair(u,v)]==2)
			Q.push(make_pair(u,v));
	}
	for(int i=1;i<=n;i++)
		f[i]=i;
	while(!Q.empty())
	{
		int u=getf(Q.front().first),v=getf(Q.front().second);
		Q.pop();
		mat[u].erase(v);mat[v].erase(u);
		if(mat[u].size()<mat[v].size())
			swap(u,v);
		for(int x:mat[v])
		{
			if(x>v)
				occur[make_pair(v,x)]=0;
			else
				occur[make_pair(x,v)]=0;
			mat[x].erase(v);
			mat[x].insert(u);
			if(x>u)
			{
				if(++occur[make_pair(u,x)]==2)
					Q.push(make_pair(u,x));
			}
			else
			{
				if(++occur[make_pair(x,u)]==2)
					Q.push(make_pair(x,u));
			}
			mat[u].insert(x);
		}
		f[v]=u;
		mat[v].clear();
	}
	int cnt=0;
	for(int i=1;i<=n;i++)
		cnt+=getf(i)==i;
	if(cnt==1)
		cout<<"YES\n";
	else
		cout<<"NO\n";
	return 0;
}