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 pi pair<int,int>
int fa[500050];
set<int> se[100010];
map<pi,int> mp;
queue<pi> q;
int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}
int main(){
	int n; cin>>n;
	for(int i=1;i<=n*2-2;i++){
		int x,y; scanf("%d%d",&x,&y);
		se[x].insert(y); se[y].insert(x);
		mp[make_pair(x,y)]++; mp[make_pair(y,x)]++;
		if(mp[make_pair(x,y)]==2) q.push(make_pair(x,y));
	}
	for(int i=1;i<=n;i++) fa[i]=i;
	set<int>::iterator it;
	while(!q.empty()){
		int x=q.front().first; int y=q.front().second; q.pop();
		x=find(x); y=find(y);
		se[x].erase(y); se[y].erase(x);
		if(se[x].size()<se[y].size()) swap(x,y);
		for(it=se[y].begin();it!=se[y].end();it++){
			int v=*it;
			mp[make_pair(v,y)]=mp[make_pair(y,v)]=0;
			se[v].erase(y); 
			se[v].insert(x);  se[x].insert(v);
			mp[make_pair(v,x)]++; mp[make_pair(x,v)]++;
			if(mp[make_pair(v,x)]==2) q.push(make_pair(v,x));
		}
		fa[y]=x; se[y].clear();
	}
	int flag=0;
	for(int i=1;i<=n;i++){
		if(fa[i]==i&&flag){
			return puts("NO"),0;
		}
		if(fa[i]==i) flag=1;
	}
	puts("YES");
}