#include<bits/stdc++.h>
using namespace std;
#define N 111116
typedef pair<int,int> edge;

int n,f[N];
multiset<int> s[N];
map<edge,int> h;
queue<edge> q;

edge make(int x,int y){
	if (x>y) swap(x,y);
	return edge(x,y);
}

int find(int x){
	if (f[x]==x) return x;
	return f[x]=find(f[x]);
}

int read(){
	int x=0,f=1;char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-')  f=-f;
	for (;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
	return x*f;
}

int main(){
	n=read();
	for (int i=1;i<=2*n-2;i++){
		int x=read(),y=read();edge e=make(x,y);
		s[x].insert(y);s[y].insert(x);
		if (++h[e]==2) q.push(e);
	}
	for (int i=1;i<=n;i++) f[i]=i;
	for (int i=1;i<=n-1;i++){
		int x=0,y=0;
		while (x==y){
			if (q.empty()) {printf("NO\n");return 0;};
			x=find(q.front().first);y=find(q.front().second);q.pop();
		}
		if (s[x].size()>s[y].size()) swap(x,y);f[x]=y;
		for (set<int>::iterator i=s[x].begin();i!=s[x].end();i++){
			int z=find(*i);edge e=make(y,z);
			if (y==z) continue;
			s[y].insert(z);s[z].insert(y);
			if (++h[e]==2) q.push(e);
			s[z].erase(s[z].find(x));
		}
		s[y].erase(s[y].find(x));s[x].clear();
	}
	printf("YES\n");return 0;
}
### Prompt

Please provide a CPP coded solution 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>
using namespace std;
#define N 111116
typedef pair<int,int> edge;

int n,f[N];
multiset<int> s[N];
map<edge,int> h;
queue<edge> q;

edge make(int x,int y){
	if (x>y) swap(x,y);
	return edge(x,y);
}

int find(int x){
	if (f[x]==x) return x;
	return f[x]=find(f[x]);
}

int read(){
	int x=0,f=1;char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-')  f=-f;
	for (;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
	return x*f;
}

int main(){
	n=read();
	for (int i=1;i<=2*n-2;i++){
		int x=read(),y=read();edge e=make(x,y);
		s[x].insert(y);s[y].insert(x);
		if (++h[e]==2) q.push(e);
	}
	for (int i=1;i<=n;i++) f[i]=i;
	for (int i=1;i<=n-1;i++){
		int x=0,y=0;
		while (x==y){
			if (q.empty()) {printf("NO\n");return 0;};
			x=find(q.front().first);y=find(q.front().second);q.pop();
		}
		if (s[x].size()>s[y].size()) swap(x,y);f[x]=y;
		for (set<int>::iterator i=s[x].begin();i!=s[x].end();i++){
			int z=find(*i);edge e=make(y,z);
			if (y==z) continue;
			s[y].insert(z);s[z].insert(y);
			if (++h[e]==2) q.push(e);
			s[z].erase(s[z].find(x));
		}
		s[y].erase(s[y].find(x));s[x].clear();
	}
	printf("YES\n");return 0;
}
```