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;
int n,fa[120000];
set <int> st[120000];
map <pair <int,int> ,int> mp;
queue <pair <int,int> > q;
int find(int x) {if(fa[x]==x) return x;return fa[x]=find(fa[x]);}
signed main() {
	ios::sync_with_stdio(false);
	cin>>n;
	for(int i=1;i<=n*2-2;i++) {
		int x,y;
		cin>>x>>y;
		st[x].insert(y);st[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;
	while(!q.empty()) {
		int x=q.front().first,y=q.front().second;
		q.pop();
		x=find(x);y=find(y);
		st[x].erase(y);
		st[y].erase(x);
		if(st[x].size()<st[y].size()) swap(x,y);
		for(set <int> :: iterator it=st[y].begin();it!=st[y].end();it++) {
			int now=*it;
			mp[make_pair(now,y)]=mp[make_pair(y,now)]=0;
			st[now].erase(y);
			st[now].insert(x);
			mp[make_pair(now,x)]++;
			mp[make_pair(x,now)]++;
			if(mp[make_pair(now,x)]==2) q.push(make_pair(now,x));
			st[x].insert(now);
		}
		fa[y]=x;
		st[y].clear();
	}
	int s=0;
	for(int i=1;i<=n;i++) if(fa[i]==i) s++;
	if(s==1) cout<<"YES";
	else cout<<"NO";
	return 0;
}