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<stdio.h>
#include<set>
using namespace std;
struct pr{
	int l,r;
	pr(int a=0,int b=0){
		if(a>b)swap(a,b);
		l=a;
		r=b;
	}
};
bool operator<(pr a,pr b){return a.l==b.l?a.r<b.r:a.l<b.l;}
set<pr>s;
set<int>g[100010];
int fa[100010];
int get(int x){
	return fa[x]==x?x:(fa[x]=get(fa[x]));
}
void add(int x,int y){
	set<int>::iterator it=g[x].find(y);
	if(it==g[x].end())
		g[x].insert(y);
	else{
		g[x].erase(it);
		s.insert(pr(x,y));
	}
}
int main(){
	int n,m,i,x,y;
	scanf("%d",&n);
	for(i=1;i<=2*n-2;i++){
		scanf("%d%d",&x,&y);
		add(x,y);
		add(y,x);
	}
	for(i=1;i<=n;i++)fa[i]=i;
	m=0;
	while(!s.empty()){
		set<pr>::iterator it=s.begin();
		x=get(it->l);
		y=get(it->r);
		s.erase(it);
		if(g[x].size()>g[y].size())swap(x,y);
		fa[x]=y;
		for(int i:g[x]){
			g[i].erase(x);
			add(i,y);
			add(y,i);
		}
		g[x].clear();
		m++;
	}
	puts(m==n-1?"YES":"NO");
}