#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
using namespace std;
inline int read(){
	int f=1,ans=0;char c=getchar();
	while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
	while(c>='0'&&c<='9'){ans=ans*10+c-'0';c=getchar();}
	return f*ans;
}
const int MAXN=200001;
map<pair<int,int> ,int> M;
set<int> s[MAXN];
queue<pair<int,int> > que;
void add(int u,int v){
	s[u].insert(v),s[v].insert(u);
	if(u>v) swap(u,v);
	M[make_pair(u,v)]++;
	if(M[make_pair(u,v)]==2) que.push(make_pair(u,v));
	return; 
}
void del(int u,int v){
	s[u].erase(v),s[v].erase(u);
	if(u>v) swap(u,v);
	M.erase(make_pair(u,v));
}
set<int>::iterator it,it1;
int n,f[MAXN];
int find(int x){if(f[x]==x) return x;return f[x]=find(f[x]);}
bool solve(){
	int tot=0;
	for(int i=1;i<=n;i++) f[i]=i;
	while(!que.empty()){
		int u=que.front().first,v=que.front().second;que.pop();
		tot++;
		u=find(u),v=find(v);
		if(s[u].size()<s[v].size()) swap(u,v);
		del(u,v);
		f[v]=u;
		for(it=s[v].begin();it!=s[v].end();){
			it1=it;it++;
			int end=*it1;end=find(end);
			del(v,end),add(u,end);
		}
//		s[v].clear();
	}
	return tot==(n-1);
}
int main(){
// 	freopen("tree.in","r",stdin);
// 	freopen("tree.out","w",stdout);
	n=read();
	for(int i=1;i<n;i++) add(read(),read());
	for(int i=1;i<n;i++) add(read(),read());
	if(solve()){printf("YES\n");return 0;}
	printf("NO\n");return 0;
}

### Prompt

In cpp, your task is to solve the following 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

### Response

```cpp
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
using namespace std;
inline int read(){
	int f=1,ans=0;char c=getchar();
	while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
	while(c>='0'&&c<='9'){ans=ans*10+c-'0';c=getchar();}
	return f*ans;
}
const int MAXN=200001;
map<pair<int,int> ,int> M;
set<int> s[MAXN];
queue<pair<int,int> > que;
void add(int u,int v){
	s[u].insert(v),s[v].insert(u);
	if(u>v) swap(u,v);
	M[make_pair(u,v)]++;
	if(M[make_pair(u,v)]==2) que.push(make_pair(u,v));
	return; 
}
void del(int u,int v){
	s[u].erase(v),s[v].erase(u);
	if(u>v) swap(u,v);
	M.erase(make_pair(u,v));
}
set<int>::iterator it,it1;
int n,f[MAXN];
int find(int x){if(f[x]==x) return x;return f[x]=find(f[x]);}
bool solve(){
	int tot=0;
	for(int i=1;i<=n;i++) f[i]=i;
	while(!que.empty()){
		int u=que.front().first,v=que.front().second;que.pop();
		tot++;
		u=find(u),v=find(v);
		if(s[u].size()<s[v].size()) swap(u,v);
		del(u,v);
		f[v]=u;
		for(it=s[v].begin();it!=s[v].end();){
			it1=it;it++;
			int end=*it1;end=find(end);
			del(v,end),add(u,end);
		}
//		s[v].clear();
	}
	return tot==(n-1);
}
int main(){
// 	freopen("tree.in","r",stdin);
// 	freopen("tree.out","w",stdout);
	n=read();
	for(int i=1;i<n;i++) add(read(),read());
	for(int i=1;i<n;i++) add(read(),read());
	if(solve()){printf("YES\n");return 0;}
	printf("NO\n");return 0;
}

```