#include<bits/stdc++.h>
#define re register int
using namespace std;
typedef pair<int,int> pir;
const int N=1e5+10;
int n;
int fa[N];
set<int> e[N];
map<pir,int> cnt;
queue<pir>q;
inline int read(){
	int x=0,f=0,ch=getchar();
    while(!isdigit(ch)){f|=ch=='-';ch=getchar();};
    while(isdigit(ch)){x=(x<<1)+(x<<3)+(ch^48),ch=getchar();}
    return f?-x:x;
}
inline int find(int x){
	return x==fa[x]?x:fa[x]=find(fa[x]);	
}
inline void add_edge(int u,int v){
	e[u].insert(v);e[v].insert(u);
	cnt[pir(u,v)]++;cnt[pir(v,u)]++;
	if(cnt[pir(u,v)]==2) q.push(pir(u,v));
}
void init(){
	scanf("%d",&n);
	for(re i=1;i<=n;i++) fa[i]=i;
	for(re i=1,u,v;i<=(n-1)<<1;i++) add_edge(u=read(),v=read());
}
void solve(){
	while(!q.empty()){
		int x=q.front().first,y=q.front().second;
		q.pop();
		x=find(x),y=find(y);
		e[x].erase(y),e[y].erase(x);
		if(e[x].size()<e[y].size()) swap(x,y);
		fa[y]=x;
		for(set<int>::iterator i=e[y].begin();i!=e[y].end();i++){
			e[*i].erase(y),e[*i].insert(x),e[x].insert(*i);
			cnt[pir(y,*i)]=cnt[pir(*i,y)]=0;
			cnt[pir(x,*i)]++;cnt[pir(*i,x)]++;
			if(cnt[pir(x,*i)]==2) q.push(pir(x,*i));
		}
		e[y].clear();
	}
}
void print(){
	int ans=0;
	for(re i=1;i<=n;i++)
		if(i==find(i))
			ans++;
	puts(ans==1?"YES":"NO");
}
int main(){
	init();
	solve();
	print();
}
### Prompt

Your challenge is to write a Cpp solution to 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<bits/stdc++.h>
#define re register int
using namespace std;
typedef pair<int,int> pir;
const int N=1e5+10;
int n;
int fa[N];
set<int> e[N];
map<pir,int> cnt;
queue<pir>q;
inline int read(){
	int x=0,f=0,ch=getchar();
    while(!isdigit(ch)){f|=ch=='-';ch=getchar();};
    while(isdigit(ch)){x=(x<<1)+(x<<3)+(ch^48),ch=getchar();}
    return f?-x:x;
}
inline int find(int x){
	return x==fa[x]?x:fa[x]=find(fa[x]);	
}
inline void add_edge(int u,int v){
	e[u].insert(v);e[v].insert(u);
	cnt[pir(u,v)]++;cnt[pir(v,u)]++;
	if(cnt[pir(u,v)]==2) q.push(pir(u,v));
}
void init(){
	scanf("%d",&n);
	for(re i=1;i<=n;i++) fa[i]=i;
	for(re i=1,u,v;i<=(n-1)<<1;i++) add_edge(u=read(),v=read());
}
void solve(){
	while(!q.empty()){
		int x=q.front().first,y=q.front().second;
		q.pop();
		x=find(x),y=find(y);
		e[x].erase(y),e[y].erase(x);
		if(e[x].size()<e[y].size()) swap(x,y);
		fa[y]=x;
		for(set<int>::iterator i=e[y].begin();i!=e[y].end();i++){
			e[*i].erase(y),e[*i].insert(x),e[x].insert(*i);
			cnt[pir(y,*i)]=cnt[pir(*i,y)]=0;
			cnt[pir(x,*i)]++;cnt[pir(*i,x)]++;
			if(cnt[pir(x,*i)]==2) q.push(pir(x,*i));
		}
		e[y].clear();
	}
}
void print(){
	int ans=0;
	for(re i=1;i<=n;i++)
		if(i==find(i))
			ans++;
	puts(ans==1?"YES":"NO");
}
int main(){
	init();
	solve();
	print();
}
```