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;

const int N=200100;
multiset<int>s[N];
map<pair<int,int>,int>mp;
queue<pair<int,int> >q;
int n;

void add(int x,int y){
	if(x>y)swap(x,y);
	if(x==y)return;
	if(++mp[make_pair(x,y)]==2)q.emplace(x,y);
	s[x].insert(y);s[y].insert(x);
}

int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n;
	for(int x,y;cin>>x>>y;add(x,y));
	for(int i=1;i<n;++i)
		for(;;){
			if(q.empty()){cout<<"NO\n";exit(0);}
			auto x=q.front();q.pop();
			if(!mp[x])continue;
			int u=x.first,v=x.second;
			if(s[u].size()<s[v].size())swap(u,v);
			for(int i:s[v]){
				mp[make_pair(min(i,v),max(i,v))]=0;
				s[i].erase(s[i].find(v));add(i,u);
			}
			s[v].clear();
			break;
		}
	cout<<"YES\n";
	return 0;
}