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 mpr make_pair
typedef pair < int , int > pii;
int n,fa[100005],cnt;
multiset < int > st[100005],con[100005];
map < int , int > ect[100005];
vector < pii > q;
int anscestor(int x){
	if(fa[x]!=x) fa[x]=anscestor(fa[x]);
	return fa[x];
}
int main(){
	scanf("%d",&n);
	int xb,x2,u,v;
	for(xb=0;xb<n*2-2;++xb){
		scanf("%d%d",&u,&v);--u;--v;
		st[u].insert(v);
		st[v].insert(u);
		++ect[u][v];
		++ect[v][u];
		if(ect[u][v]==2) q.push_back(mpr(u,v));
		fa[u]=u;
		fa[v]=v;
	}
	for(xb=0;xb<n;++xb) con[xb].insert(xb);
	set < int >::iterator it;
	map < int , int >::iterator it2;
	while(!q.empty()){
		u=q.back().first;
		v=q.back().second;
		q.pop_back();
		u=anscestor(u);
		v=anscestor(v);
		if(u==v) continue;
		if(con[u].size()+st[u].size()>con[v].size()+st[v].size()) swap(u,v);
		for(it=st[u].begin();it!=st[u].end();++it){
			if(con[v].find(*it)==con[v].end()) st[v].insert(*it);
		}
		for(it=con[u].begin();it!=con[u].end();++it){
			con[v].insert(*it);
		}
		for(it2=ect[u].begin();it2!=ect[u].end();++it2){
			ect[v][it2->first]+=it2->second;
			ect[it2->first][v]+=it2->second;
			if(ect[v][it2->first]==2) q.push_back(mpr(v,it2->first));
		}
		fa[u]=v;
		++cnt;
	}
	if(cnt==n-1) printf("YES"); else printf("NO");
	return 0;
}