#include <bits/stdc++.h>
#define Inf 0x3f3f3f3f
#define rg register
using namespace std;
typedef long long ll;
inline ll min(ll x, ll y) { return x < y ? x : y; }
inline ll max(ll x, ll y) { return x > y ? x : y; }
const int Maxn = 1e5 + 5;
inline ll read()
{
	rg ll res = 0;
	rg char ch = getchar();
	while (!isdigit(ch))
		ch = getchar();
	do
	{
		res = (res << 3) + (res << 1) + (ch ^ 48);
		ch = getchar();
	}while(isdigit(ch));
	return res;
}
int n,f[Maxn];
set<int>E[Maxn];
set<int>::iterator it;
queue< pair<int,int> >q;
int Find(int x){return f[x]==x?x:f[x]=Find(f[x]);}
int main()
{
	rg int a,b;
	cin>>n;
	for(int t=0;t<=1;t++)
		for(int i=1;i<n;i++)
		{
			a=read();
			b=read();
			if(!t)
			{
				E[a].insert(b);
				E[b].insert(a);
			}
			else
			{
				if(E[a].find(b)!=E[a].end())
					q.push(make_pair(a,b));
				E[a].insert(b);
				E[b].insert(a);
			}
		}
	for(int i=1;i<=n;i++)
		f[i]=i;
	rg int ans=n;
	while(!q.empty())
	{
		a=Find(q.front().first);
		b=Find(q.front().second);
		ans--;
		q.pop();
		if(E[a].size()<E[b].size())
			swap(a,b);
		E[b].erase(a);
		for(it=E[b].begin();it!=E[b].end();++it)
		{
			if(E[a].find(*it)!=E[a].end())
				q.push(make_pair(a,*it));
			E[a].insert(*it);
			E[*it].erase(b);
			E[*it].insert(a);
		}
		f[b]=a;
	}
	if(ans==1)
		puts("YES");
	else
		puts("NO");
	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>
#define Inf 0x3f3f3f3f
#define rg register
using namespace std;
typedef long long ll;
inline ll min(ll x, ll y) { return x < y ? x : y; }
inline ll max(ll x, ll y) { return x > y ? x : y; }
const int Maxn = 1e5 + 5;
inline ll read()
{
	rg ll res = 0;
	rg char ch = getchar();
	while (!isdigit(ch))
		ch = getchar();
	do
	{
		res = (res << 3) + (res << 1) + (ch ^ 48);
		ch = getchar();
	}while(isdigit(ch));
	return res;
}
int n,f[Maxn];
set<int>E[Maxn];
set<int>::iterator it;
queue< pair<int,int> >q;
int Find(int x){return f[x]==x?x:f[x]=Find(f[x]);}
int main()
{
	rg int a,b;
	cin>>n;
	for(int t=0;t<=1;t++)
		for(int i=1;i<n;i++)
		{
			a=read();
			b=read();
			if(!t)
			{
				E[a].insert(b);
				E[b].insert(a);
			}
			else
			{
				if(E[a].find(b)!=E[a].end())
					q.push(make_pair(a,b));
				E[a].insert(b);
				E[b].insert(a);
			}
		}
	for(int i=1;i<=n;i++)
		f[i]=i;
	rg int ans=n;
	while(!q.empty())
	{
		a=Find(q.front().first);
		b=Find(q.front().second);
		ans--;
		q.pop();
		if(E[a].size()<E[b].size())
			swap(a,b);
		E[b].erase(a);
		for(it=E[b].begin();it!=E[b].end();++it)
		{
			if(E[a].find(*it)!=E[a].end())
				q.push(make_pair(a,*it));
			E[a].insert(*it);
			E[*it].erase(b);
			E[*it].insert(a);
		}
		f[b]=a;
	}
	if(ans==1)
		puts("YES");
	else
		puts("NO");
	return 0;
}

```