#include <bits/stdc++.h>
#define rep(i,n) for ((i)=1;(i)<=(n);(i)++)
#define per(i,n) for ((i)=n;(i)>=(1);(i)--)
using namespace std;
int n,m,i,j,cnt,fa[100005];
set<int> adj[100005];
queue<pair<int,int> > q;
int fnd(int x){
	if(fa[x]==x) return x;
	return fa[x]=fnd(fa[x]); 
}
int main(){
	scanf("%d",&n);
	rep(i,n) fa[i]=i;
	rep(i,n+n-2){
		int a,b;
		scanf("%d%d",&a,&b);
		if(adj[a].count(b)) q.push(make_pair(a,b));
		adj[a].insert(b);
		adj[b].insert(a);
	}
	while(!q.empty()){
		int x=q.front().first,y=q.front().second;q.pop();
		x=fnd(x);y=fnd(y);if(x==y) continue;
		if(adj[x].size()>adj[y].size()) swap(x,y);
		cnt++;fa[x]=y;adj[x].erase(y);adj[y].erase(x);
		for(set<int>::iterator it=adj[x].begin();it!=adj[x].end();it++){
			int t=fnd(*it);if(t==y) continue;
			adj[t].erase(x);
			if(adj[y].count(t)){
				q.push(make_pair(min(t,y),max(t,y)));
			}
			adj[y].insert(t);adj[t].insert(y);
		}
		adj[x].clear();
	}
	if(cnt>=n-1) puts("YES"); else puts("NO");
	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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 rep(i,n) for ((i)=1;(i)<=(n);(i)++)
#define per(i,n) for ((i)=n;(i)>=(1);(i)--)
using namespace std;
int n,m,i,j,cnt,fa[100005];
set<int> adj[100005];
queue<pair<int,int> > q;
int fnd(int x){
	if(fa[x]==x) return x;
	return fa[x]=fnd(fa[x]); 
}
int main(){
	scanf("%d",&n);
	rep(i,n) fa[i]=i;
	rep(i,n+n-2){
		int a,b;
		scanf("%d%d",&a,&b);
		if(adj[a].count(b)) q.push(make_pair(a,b));
		adj[a].insert(b);
		adj[b].insert(a);
	}
	while(!q.empty()){
		int x=q.front().first,y=q.front().second;q.pop();
		x=fnd(x);y=fnd(y);if(x==y) continue;
		if(adj[x].size()>adj[y].size()) swap(x,y);
		cnt++;fa[x]=y;adj[x].erase(y);adj[y].erase(x);
		for(set<int>::iterator it=adj[x].begin();it!=adj[x].end();it++){
			int t=fnd(*it);if(t==y) continue;
			adj[t].erase(x);
			if(adj[y].count(t)){
				q.push(make_pair(min(t,y),max(t,y)));
			}
			adj[y].insert(t);adj[t].insert(y);
		}
		adj[x].clear();
	}
	if(cnt>=n-1) puts("YES"); else puts("NO");
	return 0;
}
```