#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<ctime>
#include<set>
#include<climits>
#include<algorithm>
#include<queue>
using namespace std;
const int N=100010;
typedef set<int>::iterator It;
set<int> s[N];
queue<pair<int,int> > q;
int n,f[N];
int find(int x) {return f[x]==x?x:f[x]=find(f[x]);}
void insert(int x,int y)
{
	if(s[x].find(y)!=s[x].end()) q.push(make_pair(x,y));
	else
	{
		s[x].insert(y);
	    s[y].insert(x);
	}
}
int main()
{
	int i,j,x,y;
	scanf("%d",&n);
	for(i=1;i<=n;i++) f[i]=i;
	for(i=1;i<n;i++)
	{
	    scanf("%d%d",&x,&y);
		s[x].insert(y);
		s[y].insert(x);
	}
	for(i=1;i<n;i++)
	{
		scanf("%d%d",&x,&y);
		insert(x,y);
	}
	for(i=1;i<n;i++)
	{
		if(q.empty()) break;
		x=q.front().first,y=q.front().second;
		q.pop();
		x=find(x),y=find(y);
		if(s[x].size()<s[y].size()) swap(x,y);
		f[y]=x;
		s[x].erase(y);
		s[y].erase(x);
		for(It it=s[y].begin();it!=s[y].end();it++) 
		{
			insert(x,find(*it));
			s[find(*it)].erase(y);
		}
	}
	if(i==n) printf("YES");
	else printf("NO");
	return 0;
}
### Prompt

Create a solution in Cpp for 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<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<ctime>
#include<set>
#include<climits>
#include<algorithm>
#include<queue>
using namespace std;
const int N=100010;
typedef set<int>::iterator It;
set<int> s[N];
queue<pair<int,int> > q;
int n,f[N];
int find(int x) {return f[x]==x?x:f[x]=find(f[x]);}
void insert(int x,int y)
{
	if(s[x].find(y)!=s[x].end()) q.push(make_pair(x,y));
	else
	{
		s[x].insert(y);
	    s[y].insert(x);
	}
}
int main()
{
	int i,j,x,y;
	scanf("%d",&n);
	for(i=1;i<=n;i++) f[i]=i;
	for(i=1;i<n;i++)
	{
	    scanf("%d%d",&x,&y);
		s[x].insert(y);
		s[y].insert(x);
	}
	for(i=1;i<n;i++)
	{
		scanf("%d%d",&x,&y);
		insert(x,y);
	}
	for(i=1;i<n;i++)
	{
		if(q.empty()) break;
		x=q.front().first,y=q.front().second;
		q.pop();
		x=find(x),y=find(y);
		if(s[x].size()<s[y].size()) swap(x,y);
		f[y]=x;
		s[x].erase(y);
		s[y].erase(x);
		for(It it=s[y].begin();it!=s[y].end();it++) 
		{
			insert(x,find(*it));
			s[find(*it)].erase(y);
		}
	}
	if(i==n) printf("YES");
	else printf("NO");
	return 0;
}
```