//just for test
#include <cstdio>
#include <set>
#include <queue>
#include <map>
#define mp make_pair
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef set<int> si;
typedef set<int>::iterator sit;
const int N=100005;
int n;
int bl[N];
si s[N];
queue<pii> q;
map<ll,int> g;
inline void swap(int &x,int &y){
	x^=y^=x^=y;
}   
int find(int u){
	return bl[u]==u?u:(bl[u]=find(bl[u]));
}
inline ll getID(int x,int y){
	if(x>y) swap(x,y);
	return 1ll*n*y+x;
}
void addEdge(int u,int v){
	s[u].insert(v);
	s[v].insert(u);
	ll eid=getID(u,v);
	int t=g[eid]+1;
	g[eid]++;
	if(t==2)
		q.push(mp(u,v));
}
void removeEdge(int u,int v){
	s[u].erase(v);
	s[v].erase(u);
	g.erase(getID(u,v));
}
void readData(){
	scanf("%d",&n);
	int u,v;
	for(int i=1;i<=(n-1)<<1;i++){
		scanf("%d%d",&u,&v);
		addEdge(u,v);
	}
}
bool solve(){
	for(int i=1;i<=n;i++) bl[i]=i;
	for(int i=1;i<n;i++){
		int u,v;
		if(q.empty())
			return false;
		u=q.front().first;
		v=q.front().second;
		q.pop();
		u=find(u); 
		v=find(v);
		if(s[u].size()>s[v].size())
			swap(u,v);
		bl[u]=v;
		removeEdge(u,v);
		for(sit i=s[u].begin(),j;i!=s[u].end();i=j){
			j=i;
			j++;
			int x=*i;
			x=find(x);
			removeEdge(u,x);
			addEdge(v,x);
		}
	}
	return true;
}
int main(){
	//freopen("a.in","r",stdin);
	readData();
	puts(solve()?"YES":"NO");
	return 0;
}
### 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
//just for test
#include <cstdio>
#include <set>
#include <queue>
#include <map>
#define mp make_pair
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef set<int> si;
typedef set<int>::iterator sit;
const int N=100005;
int n;
int bl[N];
si s[N];
queue<pii> q;
map<ll,int> g;
inline void swap(int &x,int &y){
	x^=y^=x^=y;
}   
int find(int u){
	return bl[u]==u?u:(bl[u]=find(bl[u]));
}
inline ll getID(int x,int y){
	if(x>y) swap(x,y);
	return 1ll*n*y+x;
}
void addEdge(int u,int v){
	s[u].insert(v);
	s[v].insert(u);
	ll eid=getID(u,v);
	int t=g[eid]+1;
	g[eid]++;
	if(t==2)
		q.push(mp(u,v));
}
void removeEdge(int u,int v){
	s[u].erase(v);
	s[v].erase(u);
	g.erase(getID(u,v));
}
void readData(){
	scanf("%d",&n);
	int u,v;
	for(int i=1;i<=(n-1)<<1;i++){
		scanf("%d%d",&u,&v);
		addEdge(u,v);
	}
}
bool solve(){
	for(int i=1;i<=n;i++) bl[i]=i;
	for(int i=1;i<n;i++){
		int u,v;
		if(q.empty())
			return false;
		u=q.front().first;
		v=q.front().second;
		q.pop();
		u=find(u); 
		v=find(v);
		if(s[u].size()>s[v].size())
			swap(u,v);
		bl[u]=v;
		removeEdge(u,v);
		for(sit i=s[u].begin(),j;i!=s[u].end();i=j){
			j=i;
			j++;
			int x=*i;
			x=find(x);
			removeEdge(u,x);
			addEdge(v,x);
		}
	}
	return true;
}
int main(){
	//freopen("a.in","r",stdin);
	readData();
	puts(solve()?"YES":"NO");
	return 0;
}
```