#include<bits/stdc++.h>
using namespace std;
const int N=2e5+5;
inline int read(){
	int s=0;char ch=getchar();
	while(!isdigit(ch)) ch=getchar();
	while(isdigit(ch)) s=s*10+ch-'0',ch=getchar();
	return s;
}
typedef pair<int,int> pii;
#define fi first
#define se second
int n;
set<int> s[N];
map<pii,int> mp; 
queue<pii> q;
int fa[N];
inline int findf(int x){
	return x==fa[x]?x:fa[x]=findf(fa[x]);
}
int main(){
	n=read();
	for(int i=1;i<=2*(n-1);i++){
		int x=read(),y=read();
		if(x>y) swap(x,y);
		s[x].insert(y),s[y].insert(x);
		pii p=make_pair(x,y);
		mp[p]++;
		if(mp[p]==2) q.push(p);
	}
	for(int i=1;i<=n;i++) fa[i]=i;
	for(int i=1;i<n;i++){
		if(q.empty()) {
			puts("NO");
			return 0;
		}
		int x=0,y=0;
		while(!q.empty()&&x==y) x=findf(q.front().fi),y=findf(q.front().se),q.pop();
		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 z=*it;
			s[z].erase(x);
			if(z==y) continue;
			s[z].insert(y);s[y].insert(z);
			pii p;
			if(z<y) p=make_pair(z,y);
			else p=make_pair(y,z);
			mp[p]++;
			if(mp[p]==2) q.push(p);
		}
	}
	puts("YES");
	return 0;
}
### Prompt

Your task is to create 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=2e5+5;
inline int read(){
	int s=0;char ch=getchar();
	while(!isdigit(ch)) ch=getchar();
	while(isdigit(ch)) s=s*10+ch-'0',ch=getchar();
	return s;
}
typedef pair<int,int> pii;
#define fi first
#define se second
int n;
set<int> s[N];
map<pii,int> mp; 
queue<pii> q;
int fa[N];
inline int findf(int x){
	return x==fa[x]?x:fa[x]=findf(fa[x]);
}
int main(){
	n=read();
	for(int i=1;i<=2*(n-1);i++){
		int x=read(),y=read();
		if(x>y) swap(x,y);
		s[x].insert(y),s[y].insert(x);
		pii p=make_pair(x,y);
		mp[p]++;
		if(mp[p]==2) q.push(p);
	}
	for(int i=1;i<=n;i++) fa[i]=i;
	for(int i=1;i<n;i++){
		if(q.empty()) {
			puts("NO");
			return 0;
		}
		int x=0,y=0;
		while(!q.empty()&&x==y) x=findf(q.front().fi),y=findf(q.front().se),q.pop();
		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 z=*it;
			s[z].erase(x);
			if(z==y) continue;
			s[z].insert(y);s[y].insert(z);
			pii p;
			if(z<y) p=make_pair(z,y);
			else p=make_pair(y,z);
			mp[p]++;
			if(mp[p]==2) q.push(p);
		}
	}
	puts("YES");
	return 0;
}
```