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 <cmath>
#include <set>
#include <map>
#include <queue>
#include <algorithm>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
const int N=1e5+5;
map<int,int>mp[N];
set<int>e[N];
queue<pair<int,int> >que;
int fa[N],n;
int getfather(int &p){
	return p=fa[p]==p?p:getfather(fa[p]);
}
int main(){
	scanf("%d",&n);
	rep(i,1,2*(n-1)){
		int u,v;
		scanf("%d%d",&u,&v);
		mp[u][v]++;mp[v][u]++;
		e[u].insert(v);e[v].insert(u);
		if(mp[u][v]==2) que.push(make_pair(u,v));
	}
	rep(i,1,n) fa[i]=i;
	while(!que.empty()){
		int u=que.front().first,v=que.front().second;
		getfather(u);getfather(v);
		que.pop(); e[u].erase(v); e[v].erase(u);
		mp[u][v]--;mp[v][u]--;
		if(e[u].size()<e[v].size()) swap(u,v);
		for(set<int>::iterator it=e[v].begin();it!=e[v].end();it++){
			int t=*it;
			e[t].erase(v);e[t].insert(u);e[u].insert(t);
			mp[v][t]=mp[t][v]=0;
			mp[u][t]++;mp[t][u]++;
			if(mp[u][t]==2) que.push(make_pair(u,t));
		}
		fa[v]=u;e[v].clear();
	}
	int res=0;
	rep(i,1,n) res+=fa[i]==i;
	puts(res==1?"YES":"NO");
	return 0;
}