#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int> par;
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 fa[1000005];
map <par,int>m,a;
multiset<int>s[100005];
queue<par>q;
par makep(int x,int y){
	if (x>y) return par(y,x);
	else return par(x,y);
}
int find(int x){
	return fa[x]==x?x:find(fa[x]);
}
int main(){
	int n=read();
	for (int i=1;i<=2*n-2;i++){
		int u=read(),v=read();
		par w=makep(u,v);
		m[w]++; 
		if (m[w]>1&&a[w]==0){
			a[w]=1; q.push(w);
		}
		if (i<=n) fa[i]=i; 
		s[v].insert(u); s[u].insert(v);
	}
	for (int k=1;k<n;k++){
		int x,y;
		while (1){
			if (q.empty()){printf("NO\n"); return 0;}
			par w=q.front(); q.pop();
			x=find(w.first); y=find(w.second);
			if (x!=y) break;
		}
		if (s[x].size()>s[y].size()) swap(x,y);
		fa[x]=y;
		for (set<int>::iterator i=s[x].begin();i!=s[x].end();i++){
			int v=find(*i);
			if (v==y) continue;
			par z=makep(v,y);
			m[z]++; 
			if (m[z]>1&&a[z]==0){
				q.push(z); a[z]=1;
			}
			s[v].erase(s[v].find(x)); s[y].insert(v); s[v].insert(y);
		}
		s[y].erase(x);
		s[x].clear();
	}
	printf("YES\n");
	return 0;
}
### Prompt

Please formulate 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;
typedef pair<int,int> par;
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 fa[1000005];
map <par,int>m,a;
multiset<int>s[100005];
queue<par>q;
par makep(int x,int y){
	if (x>y) return par(y,x);
	else return par(x,y);
}
int find(int x){
	return fa[x]==x?x:find(fa[x]);
}
int main(){
	int n=read();
	for (int i=1;i<=2*n-2;i++){
		int u=read(),v=read();
		par w=makep(u,v);
		m[w]++; 
		if (m[w]>1&&a[w]==0){
			a[w]=1; q.push(w);
		}
		if (i<=n) fa[i]=i; 
		s[v].insert(u); s[u].insert(v);
	}
	for (int k=1;k<n;k++){
		int x,y;
		while (1){
			if (q.empty()){printf("NO\n"); return 0;}
			par w=q.front(); q.pop();
			x=find(w.first); y=find(w.second);
			if (x!=y) break;
		}
		if (s[x].size()>s[y].size()) swap(x,y);
		fa[x]=y;
		for (set<int>::iterator i=s[x].begin();i!=s[x].end();i++){
			int v=find(*i);
			if (v==y) continue;
			par z=makep(v,y);
			m[z]++; 
			if (m[z]>1&&a[z]==0){
				q.push(z); a[z]=1;
			}
			s[v].erase(s[v].find(x)); s[y].insert(v); s[v].insert(y);
		}
		s[y].erase(x);
		s[x].clear();
	}
	printf("YES\n");
	return 0;
}
```