#include<bits/stdc++.h>
using namespace std;
const int N=100005;
typedef pair<int,int> pii;
int read(){
	int x=0,f=1; char ch=getchar();
	while (ch<'0'||ch>'9'){if (ch=='-') f=-1; ch=getchar();}
	while (ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}
	return x*f;
}
int f[N];
queue<pii>q;
set<int>b[N],r[N];
int find(int x){
	return (f[x]==x)?x:f[x]=find(f[x]);
}
int main(){
	int n=read();
	for (int i=1;i<=n;i++) f[i]=i;
	for (int i=1;i<n;i++){
		int u=read(),v=read();
		b[u].insert(v); b[v].insert(u);
	}
	for (int i=1;i<n;i++){
		int u=read(),v=read();
		r[u].insert(v); r[v].insert(u);
		if (b[u].find(v)!=b[u].end()) q.push(make_pair(u,v));
	}
	while (!q.empty()){
		int u=q.front().first,v=q.front().second; q.pop();
		if (find(u)==find(v)) continue;
		u=find(u),v=find(v);
		if (r[u].size()+b[u].size()>r[v].size()+b[v].size()) swap(u,v);
		b[u].erase(v); b[v].erase(u);
		r[u].erase(v); r[v].erase(u);
		for (set<int>::iterator it=b[u].begin();it!=b[u].end();it++){
			int x=*it;
			b[x].erase(u); b[x].insert(v);
			b[v].insert(x);
			if (r[x].find(v)!=r[x].end()) q.push(make_pair(x,v));
		}
		for (set<int>::iterator it=r[u].begin();it!=r[u].end();it++){
			int x=*it;
			r[x].erase(u); r[x].insert(v);
			r[v].insert(x);
			if (b[x].find(v)!=b[x].end()) q.push(make_pair(x,v));
		}
		f[find(u)]=find(v);
	}
	int cnt=0;
	for (int i=1;i<=n;i++) if (find(i)==i) cnt++;
	if (cnt==1) printf("YES\n"); else printf("NO\n");
	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
#include<bits/stdc++.h>
using namespace std;
const int N=100005;
typedef pair<int,int> pii;
int read(){
	int x=0,f=1; char ch=getchar();
	while (ch<'0'||ch>'9'){if (ch=='-') f=-1; ch=getchar();}
	while (ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}
	return x*f;
}
int f[N];
queue<pii>q;
set<int>b[N],r[N];
int find(int x){
	return (f[x]==x)?x:f[x]=find(f[x]);
}
int main(){
	int n=read();
	for (int i=1;i<=n;i++) f[i]=i;
	for (int i=1;i<n;i++){
		int u=read(),v=read();
		b[u].insert(v); b[v].insert(u);
	}
	for (int i=1;i<n;i++){
		int u=read(),v=read();
		r[u].insert(v); r[v].insert(u);
		if (b[u].find(v)!=b[u].end()) q.push(make_pair(u,v));
	}
	while (!q.empty()){
		int u=q.front().first,v=q.front().second; q.pop();
		if (find(u)==find(v)) continue;
		u=find(u),v=find(v);
		if (r[u].size()+b[u].size()>r[v].size()+b[v].size()) swap(u,v);
		b[u].erase(v); b[v].erase(u);
		r[u].erase(v); r[v].erase(u);
		for (set<int>::iterator it=b[u].begin();it!=b[u].end();it++){
			int x=*it;
			b[x].erase(u); b[x].insert(v);
			b[v].insert(x);
			if (r[x].find(v)!=r[x].end()) q.push(make_pair(x,v));
		}
		for (set<int>::iterator it=r[u].begin();it!=r[u].end();it++){
			int x=*it;
			r[x].erase(u); r[x].insert(v);
			r[v].insert(x);
			if (b[x].find(v)!=b[x].end()) q.push(make_pair(x,v));
		}
		f[find(u)]=find(v);
	}
	int cnt=0;
	for (int i=1;i<=n;i++) if (find(i)==i) cnt++;
	if (cnt==1) printf("YES\n"); else printf("NO\n");
	return 0;
}
```