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 <bits/stdc++.h>
using namespace std;

#define MAXN 100010

struct Edge
{
	int u,v;
	Edge(){}
	Edge(int a,int b):u(a),v(b){}
	bool operator<(Edge e)const
	{
		return u==e.u?v<e.v:u<e.u;
	}
};

int n,fa[MAXN];
set<int> edge[MAXN];
set<Edge> s;

void proc(int u,int v)
{
	if(edge[u].find(v)!=edge[u].end())
	{
		edge[u].erase(edge[u].find(v));
		s.insert(Edge(min(u,v),max(u,v)));
	}
	else edge[u].insert(v);
}

int getfa(int x)
{
	return x==fa[x]?x:fa[x]=getfa(fa[x]);
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;++i)fa[i]=i;
	for(int i=1;i<n;++i) 
	{
		int u,v;
		scanf("%d%d",&u,&v);
		edge[u].insert(v);
		edge[v].insert(u);
	}
	for(int i=1;i<n;++i)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		proc(u,v);
		proc(v,u);
	}
	int cnt=0;
	while(!s.empty())
	{
		Edge now=*s.begin();s.erase(s.begin());
//		printf("%d %d\n",now.u,now.v);
		int u=getfa(now.u),v=getfa(now.v);
		if(edge[u].size()>edge[v].size())swap(u,v);
		fa[u]=v;
		for(auto x:edge[u])
		{
			edge[x].erase(u);
			proc(v,x);
			proc(x,v);
		}
		edge[u].clear();
		++cnt;
	}
	puts(cnt==n-1?"YES":"NO");
}