#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
using namespace std;
set<int>::iterator it;
set<int>::iterator it2;
set<int>s[100005];
queue<pair<int,int> >q;
map<long long,int>mp;
int f[100005];
int getf(int x){return (f[x]==x)?x:(f[x]=getf(f[x]));}
int n;
int main()
{
	scanf("%d",&n);
	int i;
	for(i=1;i<=n-1;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		s[x].insert(y);
		s[y].insert(x);
	}
	for(i=1;i<=n-1;i++)
    {
    	int x,y;
    	scanf("%d%d",&x,&y);
		if(s[x].find(y)!=s[x].end()) 
		    q.push(make_pair(x,y));
		else
		{
			s[x].insert(y);
		    s[y].insert(x);
		}
	}
	for(i=1;i<=n;i++)
	    f[i]=i;
	for(i=1;i<n;i++)
    {
    	int x,y;
    	if(q.empty())
    	{
    		cout<<"NO";
    	    return 0;
		}
		x=q.front().first;
		y=q.front().second;
		q.pop();
		x=getf(x);
		y=getf(y);
		if(s[x].size()>s[y].size())
		    swap(x,y);
		f[x]=y;
	    s[x].erase(y);
	    s[y].erase(x);
	    for(it=s[x].begin();it!=s[x].end();it++)
	    {
	    	int num=*it;
	    	if(num==x)
			    continue;
	        s[num].erase(x);
	        if(s[y].find(num)!=s[y].end())
		        q.push(make_pair(y,num));
			else
			{
				s[num].insert(y);
			    s[y].insert(num);
			}
		}
		s[x].clear();
	}
	cout<<"YES";
}
### Prompt

Construct a CPP code solution to the problem outlined:

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<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
using namespace std;
set<int>::iterator it;
set<int>::iterator it2;
set<int>s[100005];
queue<pair<int,int> >q;
map<long long,int>mp;
int f[100005];
int getf(int x){return (f[x]==x)?x:(f[x]=getf(f[x]));}
int n;
int main()
{
	scanf("%d",&n);
	int i;
	for(i=1;i<=n-1;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		s[x].insert(y);
		s[y].insert(x);
	}
	for(i=1;i<=n-1;i++)
    {
    	int x,y;
    	scanf("%d%d",&x,&y);
		if(s[x].find(y)!=s[x].end()) 
		    q.push(make_pair(x,y));
		else
		{
			s[x].insert(y);
		    s[y].insert(x);
		}
	}
	for(i=1;i<=n;i++)
	    f[i]=i;
	for(i=1;i<n;i++)
    {
    	int x,y;
    	if(q.empty())
    	{
    		cout<<"NO";
    	    return 0;
		}
		x=q.front().first;
		y=q.front().second;
		q.pop();
		x=getf(x);
		y=getf(y);
		if(s[x].size()>s[y].size())
		    swap(x,y);
		f[x]=y;
	    s[x].erase(y);
	    s[y].erase(x);
	    for(it=s[x].begin();it!=s[x].end();it++)
	    {
	    	int num=*it;
	    	if(num==x)
			    continue;
	        s[num].erase(x);
	        if(s[y].find(num)!=s[y].end())
		        q.push(make_pair(y,num));
			else
			{
				s[num].insert(y);
			    s[y].insert(num);
			}
		}
		s[x].clear();
	}
	cout<<"YES";
}
```