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<cstdio>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
using namespace std;
const int N=100005;
typedef pair<int,int> pi;
typedef set<int>::iterator sit;
int n,fa[N],cnt;
queue<pi> q;
set<int> G1[N],G2[N];
int find_father(int x){
	if(fa[x]==x)
        return x;
	return fa[x]=find_father(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);
		G1[u].insert(v);
		G1[v].insert(u);
	}
	for(int i=1;i<n;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		G2[u].insert(v);
		G2[v].insert(u);
		if(G1[u].find(v)!=G1[u].end())
            q.push(pi(u,v));
	}
	while(!q.empty()){
		int u=q.front().first,v=q.front().second;
		q.pop();
		u=find_father(u),v=find_father(v);
		if(u==v)
            continue;
		if(G1[u].size()+G2[u].size()>G1[v].size()+G2[v].size())
            swap(u,v);
		fa[u]=v;
		G1[u].erase(v);
		G1[v].erase(u);
		G2[u].erase(v);
		G2[v].erase(u);
		for(sit it=G1[u].begin();it!=G1[u].end();it++){
			int x=*it;
			G1[x].erase(u);
			G1[x].insert(v);
			G1[v].insert(x);
			if(G2[x].find(v)!=G2[x].end())
                q.push(pi(x,v));
		}
		for(sit it=G2[u].begin();it!=G2[u].end();it++){
			int x=*it;
			G2[x].erase(u);
			G2[x].insert(v);
			G2[v].insert(x);
			if(G1[x].find(v)!=G1[x].end())
                q.push(pi(x,v));
		}
	}
	int cnt=0;
	for(int i=1;i<=n;i++)
		if(find_father(i)==i)
            cnt++;
	puts(cnt==1?"YES":"NO");
}