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>
#define X first
#define Y second
using namespace std;
int n,p[100005];
typedef pair<int,int> pii;
map<pii,bool> ex;
set<int> s[100005];
queue<pii> q;
inline int fnd(int x){
	if(p[x]==x)return x;
	return p[x]=fnd(p[x]);
}
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>n;
	for(int i=0;i<n*2-2;i++){
		int u,v;
		cin>>u>>v;
		if(s[u].count(v))q.push(pii(u,v));
		else{
			s[u].insert(v);
			s[v].insert(u);
		}
	}
	for(int i=1;i<=n;i++)p[i]=i;
	while(!q.empty()){
		int u=fnd(q.front().X),v=fnd(q.front().Y);
		q.pop();
		if(u==v)continue;
		if(s[u].size()<s[v].size())swap(u,v);
		s[u].erase(v);
		s[v].erase(u);
		for(auto i:s[v]){
			s[i].erase(v);
			if(s[i].count(u))q.push(pii(u,i));
			else{
				s[u].insert(i);
				s[i].insert(u);
			}
		}
		p[v]=u;
		s[v].clear();
	}
	for(int i=2;i<=n;i++){
		if(fnd(i)!=fnd(1)){
			cout<<"NO\n";
			return 0;
		}
	}
	cout<<"YES\n";
	return 0;
}