#include <map>
#include <set>
#include <queue>
#include <cstdio>
#include <iostream>
using namespace std;
const int N=100005;
int n,fa[N];
inline int Find(int x){return fa[x]==x?x:fa[x]=Find(fa[x]);}
typedef pair<int,int> P;
map<P,int>mp;
queue<P>q;
set<int>s[N];
int main()
{
	int i,x,y,to,re=0; scanf("%d",&n); for(i=1;i<=n;i++)fa[i]=i;
	for(i=1;i<=n*2-2;i++)
	{
		scanf("%d%d",&x,&y); mp[P(x,y)]++; mp[P(y,x)]++; if(mp[P(x,y)]==2)q.push(P(x,y)); s[x].insert(y); s[y].insert(x);
	}set<int>::iterator it;
	while(!q.empty())
	{
		x=q.front().first; y=q.front().second; q.pop(); x=Find(x); y=Find(y); s[x].erase(y); s[y].erase(x);
		for(it=s[y].begin();it!=s[y].end();it++)
		{
			to=*it; s[to].erase(y); s[to].insert(x); s[x].insert(to); mp[P(y,to)]=mp[P(to,y)]=0; mp[P(x,to)]++; mp[P(to,x)]++; if(mp[P(x,to)]==2) q.push(P(x,to));
		}fa[y]=x; s[y].clear();
	}for(i=1;i<=n;i++)re+=(fa[i]==i); if(re==1)printf("YES\n");else printf("NO\n");
}
### 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 <map>
#include <set>
#include <queue>
#include <cstdio>
#include <iostream>
using namespace std;
const int N=100005;
int n,fa[N];
inline int Find(int x){return fa[x]==x?x:fa[x]=Find(fa[x]);}
typedef pair<int,int> P;
map<P,int>mp;
queue<P>q;
set<int>s[N];
int main()
{
	int i,x,y,to,re=0; scanf("%d",&n); for(i=1;i<=n;i++)fa[i]=i;
	for(i=1;i<=n*2-2;i++)
	{
		scanf("%d%d",&x,&y); mp[P(x,y)]++; mp[P(y,x)]++; if(mp[P(x,y)]==2)q.push(P(x,y)); s[x].insert(y); s[y].insert(x);
	}set<int>::iterator it;
	while(!q.empty())
	{
		x=q.front().first; y=q.front().second; q.pop(); x=Find(x); y=Find(y); s[x].erase(y); s[y].erase(x);
		for(it=s[y].begin();it!=s[y].end();it++)
		{
			to=*it; s[to].erase(y); s[to].insert(x); s[x].insert(to); mp[P(y,to)]=mp[P(to,y)]=0; mp[P(x,to)]++; mp[P(to,x)]++; if(mp[P(x,to)]==2) q.push(P(x,to));
		}fa[y]=x; s[y].clear();
	}for(i=1;i<=n;i++)re+=(fa[i]==i); if(re==1)printf("YES\n");else printf("NO\n");
}
```