#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<set>
#include<queue>
#include<map>
#define xx first 
#define yy second
using namespace std;
inline int read()
{
	int k=0,f=1;char c=getchar();
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) k=k*10+c-'0';return f*k;
}
typedef pair<int,int> P;
const int N=200055;
int n,fa[N];
map<P,int> mp;
multiset<int> G[N];
queue<P> q;
int find(int x)
{
	return fa[x]==x?x:fa[x]=find(fa[x]);
}
void add(int a,int b)
{
	G[a].insert(b);
	G[b].insert(a);
}
int main()
{
	n=read();
	for(int i=1;i<=n;i++)
		fa[i]=i;
	for(int i=1;i<=(n-1)<<1;i++)
	{
		int a,b;
		a=read();b=read();
		if(a>b) swap(a,b);
		add(a,b);
		P e=P(a,b);
		mp[e]++;
		if(mp[e]==2) 
			q.push(e);
		else if(mp[e]>2)
			{puts("NO");return 0;}
	}
	for(int i=1;i<n;i++)
	{
		if(q.empty()) {puts("NO");return 0;}
		int x=0,y=0;
		for(;x==y;x=find(q.front().xx),y=find(q.front().yy),q.pop());
		if(G[x].size()>G[y].size()) swap(x,y);
		fa[x]=y;
		for(multiset<int>::iterator it=G[x].begin();it!=G[x].end();it++)
		{
			int v=find(*it);	
			G[v].erase(G[v].find(x));
			if(v==y) continue;
			add(v,y);
			P e=P(v,y);
			if(v>y) swap(e.xx,e.yy);
			mp[e]++;
			if(mp[e]==2) q.push(e);
		}
		G[x].clear();
	}
	puts("YES");
	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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<set>
#include<queue>
#include<map>
#define xx first 
#define yy second
using namespace std;
inline int read()
{
	int k=0,f=1;char c=getchar();
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) k=k*10+c-'0';return f*k;
}
typedef pair<int,int> P;
const int N=200055;
int n,fa[N];
map<P,int> mp;
multiset<int> G[N];
queue<P> q;
int find(int x)
{
	return fa[x]==x?x:fa[x]=find(fa[x]);
}
void add(int a,int b)
{
	G[a].insert(b);
	G[b].insert(a);
}
int main()
{
	n=read();
	for(int i=1;i<=n;i++)
		fa[i]=i;
	for(int i=1;i<=(n-1)<<1;i++)
	{
		int a,b;
		a=read();b=read();
		if(a>b) swap(a,b);
		add(a,b);
		P e=P(a,b);
		mp[e]++;
		if(mp[e]==2) 
			q.push(e);
		else if(mp[e]>2)
			{puts("NO");return 0;}
	}
	for(int i=1;i<n;i++)
	{
		if(q.empty()) {puts("NO");return 0;}
		int x=0,y=0;
		for(;x==y;x=find(q.front().xx),y=find(q.front().yy),q.pop());
		if(G[x].size()>G[y].size()) swap(x,y);
		fa[x]=y;
		for(multiset<int>::iterator it=G[x].begin();it!=G[x].end();it++)
		{
			int v=find(*it);	
			G[v].erase(G[v].find(x));
			if(v==y) continue;
			add(v,y);
			P e=P(v,y);
			if(v>y) swap(e.xx,e.yy);
			mp[e]++;
			if(mp[e]==2) q.push(e);
		}
		G[x].clear();
	}
	puts("YES");
	return 0;
}
```