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 pi pair<int,int>
#define mp(x,y) make_pair(x,y)
using namespace std;
inline int read(){
	int x=0;char c=getchar();bool p=1;
	for(;!isdigit(c);c=getchar())if(c=='-')p=0;
	for(;isdigit(c);c=getchar())x=(x<<1)+(x<<3)+(c^48);
	return p?x:-x;
}
const int N=100005;
map<pi,int> p;
queue<pi> q;
set<int> s[N];
int n,x,y,t,fa[N];
int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}
int main(){
	n=read();
	for(int i=1;i<=n;i++)fa[i]=i;
	for(int i=1;i<n*2-1;i++){
		x=read();y=read();
		s[x].insert(y);s[y].insert(x);
		p[mp(x,y)]++;p[mp(y,x)]++;
		if(p[mp(x,y)]==2)q.push(mp(x,y));
	}
	set<int>::iterator it;
	while(!q.empty()){
		pi u=q.front();q.pop();
		x=find(u.first);y=find(u.second);
		s[x].erase(y);s[y].erase(x);
		if(s[x].size()<s[y].size())swap(x,y);
		for(it=s[y].begin();it!=s[y].end();it++){
			t=*it;
			p[mp(t,y)]=p[mp(y,t)]=0;
			s[t].erase(y);s[t].insert(x);
			p[mp(t,x)]++;p[mp(x,t)]++;
			if(p[mp(t,x)]==2)q.push(mp(t,x));
			s[x].insert(t);
		}
		fa[y]=x;
		s[y].clear();
	}
	int s=0;
	for(int i=1;i<=n;i++)s+=fa[i]==i;
	if(s==1)puts("YES");else puts("NO");
	return 0;
}