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 maxn 100005
using namespace std;

int n;
map<pair<int,int>,int >M;
set<pair<int,int> >st;
multiset<int>ed[maxn],ned;

void add(int u,int v){
	if(u > v) swap(u,v);
	ed[u].insert(v);
	ed[v].insert(u);
	if(++M[make_pair(u,v)] == 2)
		st.insert(make_pair(u,v));
}
void del(int u,int v){
	if(u > v) swap(u,v);
	ed[u].erase(ed[u].find(v));
	ed[v].erase(ed[v].find(u));
	if(--M[make_pair(u,v)] == 1)
		st.erase(make_pair(u,v));
}

int main(){
	scanf("%d",&n);
	for(int i=1;i<n;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		add(u,v);
	}
	for(int i=1;i<n;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		add(u,v);
	}
	for(int i=1;i<n;i++){
		if(st.empty()){
			puts("NO");
			return 0;
		}
		pair<int,int>t = *st.begin();
		int u = t.first , v = t.second;
		if(ed[u].size() > ed[v].size()) swap(u,v);
		ned = ed[u];
		for(int p:ned) del(u,p);
		for(int p:ned) if(p != v)
			add(p,v);
	}
	puts("YES");
}