#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
#define maxn 100100
#define mp make_pair
#define x first
#define y second
typedef pair<int,int> pii;
int fa[maxn];inline int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}
set<int> s[maxn];map<pii,int> cnt;int n;queue<pii> q;
inline void init()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) fa[i]=i;
	for(int i=1,a,b;i<=((n-1)<<1);i++)
	{
		scanf("%d%d",&a,&b);s[a].insert(b);s[b].insert(a);
		cnt[mp(a,b)]++;cnt[mp(b,a)]++;
		if(cnt[mp(a,b)]==2) q.push(mp(a,b));
	}
}
int main()
{
	init();
	set<int>::iterator it;
	while(q.size())
	{
		pii a=q.front();q.pop();int x=find(a.x),y=find(a.y);
		if(s[x].size()<s[y].size()) swap(x,y);s[x].erase(y);s[y].erase(x);
		for(it=s[y].begin();it!=s[y].end();it++)
		{
			int t=*it;cnt[mp(t,y)]=cnt[mp(y,t)]=0;
			s[t].erase(y);s[t].insert(x);s[x].insert(t);
			cnt[mp(t,x)]++;cnt[mp(x,t)]++;
			if(cnt[mp(t,x)]==2) q.push(mp(t,x));
		}
		fa[y]=x;s[y].clear();
	}
	int cnt=0;
	for(int i=1;i<=n;i++) if(find(i)==i) cnt++;
	puts(cnt>1?"NO":"YES");
}
### Prompt

Your challenge is to write 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
#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
#define maxn 100100
#define mp make_pair
#define x first
#define y second
typedef pair<int,int> pii;
int fa[maxn];inline int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}
set<int> s[maxn];map<pii,int> cnt;int n;queue<pii> q;
inline void init()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) fa[i]=i;
	for(int i=1,a,b;i<=((n-1)<<1);i++)
	{
		scanf("%d%d",&a,&b);s[a].insert(b);s[b].insert(a);
		cnt[mp(a,b)]++;cnt[mp(b,a)]++;
		if(cnt[mp(a,b)]==2) q.push(mp(a,b));
	}
}
int main()
{
	init();
	set<int>::iterator it;
	while(q.size())
	{
		pii a=q.front();q.pop();int x=find(a.x),y=find(a.y);
		if(s[x].size()<s[y].size()) swap(x,y);s[x].erase(y);s[y].erase(x);
		for(it=s[y].begin();it!=s[y].end();it++)
		{
			int t=*it;cnt[mp(t,y)]=cnt[mp(y,t)]=0;
			s[t].erase(y);s[t].insert(x);s[x].insert(t);
			cnt[mp(t,x)]++;cnt[mp(x,t)]++;
			if(cnt[mp(t,x)]==2) q.push(mp(t,x));
		}
		fa[y]=x;s[y].clear();
	}
	int cnt=0;
	for(int i=1;i<=n;i++) if(find(i)==i) cnt++;
	puts(cnt>1?"NO":"YES");
}
```