//by xiqiao
#include<bits/stdc++.h>
#define maxn 100005
using namespace std;
typedef long long ll;
int n;
vector <int> v[maxn];
map <pair<int,int> ,int> mp,vis;
int fa[maxn];
queue <pair<int,int> > q;
set <int> s[maxn];
int find(int x){
	if(fa[x]==x)return x;
	return fa[x]=find(fa[x]);
}
int main(){
	//freopen("1.in","r",stdin);
	scanf("%d",&n);
	for(int i=1;i<=n;i++)fa[i]=i;
	for(int i=1;i<n+n-1;i++){
		int x,y;scanf("%d%d",&x,&y);if(x>y)swap(x,y);
		s[x].insert(y);s[y].insert(x);
		mp[make_pair(x,y)]++;
		if(mp[make_pair(x,y)]==2){
			vis[make_pair(x,y)]=1;q.push(make_pair(x,y));
		}
	}
	for(int i=1;i<n;i++){
		int x,y;
		while(1){
			if(q.empty()){
				puts("NO");return 0;
			}
			x=q.front().first,y=q.front().second;q.pop();
			x=find(x);y=find(y);//cout<<x<<" "<<y<<endl;
			if(x!=y)break;
		}
		if(s[x].size()>s[y].size())swap(x,y);
		fa[x]=y;
		for(set <int> :: iterator it=s[x].begin();it!=s[x].end();it++){
			int v=find(*it);
			if(v==y)continue;
			s[v].insert(y);s[y].insert(v);int tt=0;
			if(v>y)tt=1,swap(v,y);
			mp[make_pair(v,y)]++;
			if(mp[make_pair(v,y)]>1 && !vis[make_pair(v,y)]){
				vis[make_pair(v,y)]=1;
				q.push(make_pair(v,y));
			}
			if(tt)y=v;
		}
		s[y].erase(x);s[x].clear();
	}
	puts("YES");
	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
//by xiqiao
#include<bits/stdc++.h>
#define maxn 100005
using namespace std;
typedef long long ll;
int n;
vector <int> v[maxn];
map <pair<int,int> ,int> mp,vis;
int fa[maxn];
queue <pair<int,int> > q;
set <int> s[maxn];
int find(int x){
	if(fa[x]==x)return x;
	return fa[x]=find(fa[x]);
}
int main(){
	//freopen("1.in","r",stdin);
	scanf("%d",&n);
	for(int i=1;i<=n;i++)fa[i]=i;
	for(int i=1;i<n+n-1;i++){
		int x,y;scanf("%d%d",&x,&y);if(x>y)swap(x,y);
		s[x].insert(y);s[y].insert(x);
		mp[make_pair(x,y)]++;
		if(mp[make_pair(x,y)]==2){
			vis[make_pair(x,y)]=1;q.push(make_pair(x,y));
		}
	}
	for(int i=1;i<n;i++){
		int x,y;
		while(1){
			if(q.empty()){
				puts("NO");return 0;
			}
			x=q.front().first,y=q.front().second;q.pop();
			x=find(x);y=find(y);//cout<<x<<" "<<y<<endl;
			if(x!=y)break;
		}
		if(s[x].size()>s[y].size())swap(x,y);
		fa[x]=y;
		for(set <int> :: iterator it=s[x].begin();it!=s[x].end();it++){
			int v=find(*it);
			if(v==y)continue;
			s[v].insert(y);s[y].insert(v);int tt=0;
			if(v>y)tt=1,swap(v,y);
			mp[make_pair(v,y)]++;
			if(mp[make_pair(v,y)]>1 && !vis[make_pair(v,y)]){
				vis[make_pair(v,y)]=1;
				q.push(make_pair(v,y));
			}
			if(tt)y=v;
		}
		s[y].erase(x);s[x].clear();
	}
	puts("YES");
	return 0;
}
```