#include<bits/stdc++.h>
using namespace std;
#define debug(x) cerr<<#x<<" = "<<x
#define sp <<"  "
#define el <<endl
#define fgx cerr<<" ------------------------------------------------- "<<endl
#define LL long long
#define DB double
#define LDB long double
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
inline int read(){
	int nm=0; bool fh=true; char cw=getchar();
	for(;!isdigit(cw);cw=getchar()) fh^=(cw=='-');
	for(;isdigit(cw);cw=getchar()) nm=nm*10+(cw-'0');
	return fh?nm:-nm;
}
#define M 200020
#define IT set<int>::iterator
int n,m,f[M],u[M],v[M],hd,tl,cnt;
int tot=0;
inline LL ID(int x,int y){if(x>y)swap(x,y);return x*(LL)(n+1)+y;}
set<int>to[M]; map<LL,int>CT;
inline int fd(int x){return f[x]==x?x:f[x]=fd(f[x]);}
inline void add(int x,int y){
	if((x=fd(x))==(y=fd(y))) return;
	to[x].insert(y),to[y].insert(x);
	if(++CT[ID(x,y)]==2) u[tl]=x,v[tl++]=y;
}
inline void del(int x,int y){
	x=fd(x),y=fd(y),CT[ID(x,y)]=0;
	to[x].erase(to[x].find(y));
	to[y].erase(to[y].find(x));
}
int main(){
	n=read();
	for(int i=1;i<=n;i++) f[i]=i;
	for(int i=1,x,y;i<n;i++) x=read(),y=read(),add(x,y);
	for(int i=1,x,y;i<n;i++) x=read(),y=read(),add(x,y);
	for(int x=u[hd],y=v[hd];hd<tl;x=u[++hd],y=v[hd],++cnt){
		x=fd(x),y=fd(y); if(to[x].size()>to[y].size()) swap(x,y); del(x,y);
		for(IT it=to[x].begin();it!=to[x].end();it=to[x].begin()){
			int k=(*it); del(x,k),add(y,k);
		} f[x]=y;
	} puts((cnt<n-1)?"NO":"YES"); return 0;
}
### 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<bits/stdc++.h>
using namespace std;
#define debug(x) cerr<<#x<<" = "<<x
#define sp <<"  "
#define el <<endl
#define fgx cerr<<" ------------------------------------------------- "<<endl
#define LL long long
#define DB double
#define LDB long double
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
inline int read(){
	int nm=0; bool fh=true; char cw=getchar();
	for(;!isdigit(cw);cw=getchar()) fh^=(cw=='-');
	for(;isdigit(cw);cw=getchar()) nm=nm*10+(cw-'0');
	return fh?nm:-nm;
}
#define M 200020
#define IT set<int>::iterator
int n,m,f[M],u[M],v[M],hd,tl,cnt;
int tot=0;
inline LL ID(int x,int y){if(x>y)swap(x,y);return x*(LL)(n+1)+y;}
set<int>to[M]; map<LL,int>CT;
inline int fd(int x){return f[x]==x?x:f[x]=fd(f[x]);}
inline void add(int x,int y){
	if((x=fd(x))==(y=fd(y))) return;
	to[x].insert(y),to[y].insert(x);
	if(++CT[ID(x,y)]==2) u[tl]=x,v[tl++]=y;
}
inline void del(int x,int y){
	x=fd(x),y=fd(y),CT[ID(x,y)]=0;
	to[x].erase(to[x].find(y));
	to[y].erase(to[y].find(x));
}
int main(){
	n=read();
	for(int i=1;i<=n;i++) f[i]=i;
	for(int i=1,x,y;i<n;i++) x=read(),y=read(),add(x,y);
	for(int i=1,x,y;i<n;i++) x=read(),y=read(),add(x,y);
	for(int x=u[hd],y=v[hd];hd<tl;x=u[++hd],y=v[hd],++cnt){
		x=fd(x),y=fd(y); if(to[x].size()>to[y].size()) swap(x,y); del(x,y);
		for(IT it=to[x].begin();it!=to[x].end();it=to[x].begin()){
			int k=(*it); del(x,k),add(y,k);
		} f[x]=y;
	} puts((cnt<n-1)?"NO":"YES"); return 0;
}
```