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 N 100005
set<int>s[N];
map<int,int>mat[N];
queue<pair<int,int> >q;
int n,x,y,ans,f[N];
int find(int k){
	if (k==f[k])return k;
	return f[k]=find(f[k]);
}
void add(int x,int y){
	s[x].insert(y);
	s[y].insert(x);
	if (x>y)swap(x,y);
	if (++mat[x][y]==2){
		q.push(make_pair(x,y));
		mat[x][y]=0;
	}
}
void merge(int x,int y){
	if (s[x].size()>s[y].size())swap(x,y);
	f[x]=y;
	for(set<int>::iterator it=s[x].begin();it!=s[x].end();it++){
		int z=find(*it);
		if (y==z)continue;
		add(y,z);
		s[z].erase(x);
	}
	s[x].clear();
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)f[i]=i;
	for(int i=1;i<2*n-1;i++){
		scanf("%d%d",&x,&y);
		add(x,y);
	}
	while (!q.empty()){
		ans++;
		merge(find(q.front().first),find(q.front().second));
		q.pop();
	}
	if (ans==n-1)printf("YES");
	else printf("NO");
}