#include<bits/stdc++.h>
using namespace std;

int n,t,w,cnt;
int fa[200010],d[200010];
pair<int,int> q[4000010];
map<int,int> h[200010];

int find(int x)
{return fa[x]==x?x:fa[x]=find(fa[x]);
}

int main()
{int i,j,x,y,k;
 map<int,int>::iterator p;
 
 scanf("%d",&n);
 for(i=1;i<n;i++)
 {	scanf("%d%d",&x,&y);
 	h[x][y]=1;
 	h[y][x]=1;
 	d[x]++;
 	d[y]++;
 }
 t=1;
 w=0;
 for(i=1;i<n;i++)
 {	scanf("%d%d",&x,&y);
 	h[x][y]++;
 	h[y][x]++;
 	d[x]++;
 	d[y]++;
 	if(h[x][y]==2) q[++w]=make_pair(x,y);
 }
 for(i=1;i<=n;i++) fa[i]=i;
 while(t<=w)
 {	x=find(q[t].first);
 	y=find(q[t].second);
 	if(x!=y)
 	{	if(d[x]<d[y]) swap(x,y);
 		h[x][y]-=2;
 		h[y][x]-=2;
 		for(p=h[y].begin();p!=h[y].end();p++)
 		{	if((*p).second==0) continue;
	 		k=(h[x][(*p).first]+=(*p).second);
 			h[(*p).first][x]+=(*p).second;
 			h[(*p).first][y]-=(*p).second;
 			if(k==2)
 				q[++w]=make_pair(x,(*p).first);
 		}
 		h[y].clear();
		d[x]+=d[y]-2;
		fa[y]=x;
		cnt++;
	}
 	t++;
 }
 printf("%s\n",cnt==n-1?"YES":"NO");
 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<bits/stdc++.h>
using namespace std;

int n,t,w,cnt;
int fa[200010],d[200010];
pair<int,int> q[4000010];
map<int,int> h[200010];

int find(int x)
{return fa[x]==x?x:fa[x]=find(fa[x]);
}

int main()
{int i,j,x,y,k;
 map<int,int>::iterator p;
 
 scanf("%d",&n);
 for(i=1;i<n;i++)
 {	scanf("%d%d",&x,&y);
 	h[x][y]=1;
 	h[y][x]=1;
 	d[x]++;
 	d[y]++;
 }
 t=1;
 w=0;
 for(i=1;i<n;i++)
 {	scanf("%d%d",&x,&y);
 	h[x][y]++;
 	h[y][x]++;
 	d[x]++;
 	d[y]++;
 	if(h[x][y]==2) q[++w]=make_pair(x,y);
 }
 for(i=1;i<=n;i++) fa[i]=i;
 while(t<=w)
 {	x=find(q[t].first);
 	y=find(q[t].second);
 	if(x!=y)
 	{	if(d[x]<d[y]) swap(x,y);
 		h[x][y]-=2;
 		h[y][x]-=2;
 		for(p=h[y].begin();p!=h[y].end();p++)
 		{	if((*p).second==0) continue;
	 		k=(h[x][(*p).first]+=(*p).second);
 			h[(*p).first][x]+=(*p).second;
 			h[(*p).first][y]-=(*p).second;
 			if(k==2)
 				q[++w]=make_pair(x,(*p).first);
 		}
 		h[y].clear();
		d[x]+=d[y]-2;
		fa[y]=x;
		cnt++;
	}
 	t++;
 }
 printf("%s\n",cnt==n-1?"YES":"NO");
 return 0;
}
```