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<iostream>
#include<cstdio>
#include<set>
using namespace std;
struct data{
	int x,y;
}t[200001];
int n,m,f[200001],x,y,cnt;
set<int> v[200001];
int fa(int a){
	if(f[a]!=a)f[a]=fa(f[a]);
	return f[a];
}
void ins(int a,int b){
	set<int>::iterator id1=v[a].find(b),id2=v[b].find(a);
	if(id1==v[a].end()){
		v[a].insert(b);
		v[b].insert(a);
	}else{
		cnt++;
		v[a].erase(b);
		v[b].erase(a);
		t[cnt].x=a;
		t[cnt].y=b;
	}
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n*2-2;i++){
		scanf("%d%d",&x,&y);
		ins(x,y);
	}
	for(int i=1;i<=n;i++)f[i]=i;
	while(cnt){
		x=fa(t[cnt].x);
		y=fa(t[cnt].y);
		cnt--;
		if(v[x].size()>v[y].size())swap(x,y);
		f[x]=y;
		for(int i:v[x]){
			v[i].erase(x);
			ins(i,y);
		}
		v[x].clear();
		m++;
	}
	if(m==n-1)printf("YES\n");
	else printf("NO\n");
}