#include<bits/stdc++.h>
using namespace std;
#define mp make_pair
const int N=100001;
int fa[N],h,t,u,v,s,n,i,to;
set<int>S[N];
set<int>::iterator it;
map<pair<int,int>,int>ma;
pair<int,int>q[N];
inline char gc(){
	static char buf[100000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline int rd(){
	int x=0,fl=1;char ch=gc();
	for (;ch<48||ch>57;ch=gc())if(ch=='-')fl=-1;
	for (;48<=ch&&ch<=57;ch=gc())x=(x<<3)+(x<<1)+(ch^48);
	return x*fl;
}
inline void wri(int a){if(a<0)a=-a,putchar('-');if(a>=10)wri(a/10);putchar(a%10|48);}
inline void wln(int a){wri(a);puts("");}
int find(int x){
	return x==fa[x]?x:fa[x]=find(fa[x]);
}
int main(){
	n=rd();
	for (i=1;i<=2*n-2;i++){
		u=rd();v=rd();
		S[u].insert(v);S[v].insert(u);
		ma[mp(u,v)]++,ma[mp(v,u)]++;
		if (ma[mp(u,v)]==2) q[t++]=mp(u,v);
	}
	for (i=1;i<=n;i++) fa[i]=i;
	while (h<t){
		u=find(q[h].first),v=find(q[h++].second);
		S[u].erase(v);S[v].erase(u);
		if (S[u].size()<S[v].size()) swap(u,v);
		for (it=S[v].begin();it!=S[v].end();it++){
			to=*it;
			S[to].erase(v);
			ma[mp(v,to)]=ma[mp(to,v)]=0;
			S[u].insert(to);S[to].insert(u);
			ma[mp(u,to)]++;ma[mp(to,u)]++;
			if (ma[mp(u,to)]==2) q[t++]=mp(u,to);
		}
		fa[v]=u;S[v].clear();
	}
	for (i=1;i<=n;i++) s+=fa[i]==i;
	puts(s==1?"YES":"NO");
}
### Prompt

Develop a solution in Cpp to the problem described below:

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 mp make_pair
const int N=100001;
int fa[N],h,t,u,v,s,n,i,to;
set<int>S[N];
set<int>::iterator it;
map<pair<int,int>,int>ma;
pair<int,int>q[N];
inline char gc(){
	static char buf[100000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline int rd(){
	int x=0,fl=1;char ch=gc();
	for (;ch<48||ch>57;ch=gc())if(ch=='-')fl=-1;
	for (;48<=ch&&ch<=57;ch=gc())x=(x<<3)+(x<<1)+(ch^48);
	return x*fl;
}
inline void wri(int a){if(a<0)a=-a,putchar('-');if(a>=10)wri(a/10);putchar(a%10|48);}
inline void wln(int a){wri(a);puts("");}
int find(int x){
	return x==fa[x]?x:fa[x]=find(fa[x]);
}
int main(){
	n=rd();
	for (i=1;i<=2*n-2;i++){
		u=rd();v=rd();
		S[u].insert(v);S[v].insert(u);
		ma[mp(u,v)]++,ma[mp(v,u)]++;
		if (ma[mp(u,v)]==2) q[t++]=mp(u,v);
	}
	for (i=1;i<=n;i++) fa[i]=i;
	while (h<t){
		u=find(q[h].first),v=find(q[h++].second);
		S[u].erase(v);S[v].erase(u);
		if (S[u].size()<S[v].size()) swap(u,v);
		for (it=S[v].begin();it!=S[v].end();it++){
			to=*it;
			S[to].erase(v);
			ma[mp(v,to)]=ma[mp(to,v)]=0;
			S[u].insert(to);S[to].insert(u);
			ma[mp(u,to)]++;ma[mp(to,u)]++;
			if (ma[mp(u,to)]==2) q[t++]=mp(u,to);
		}
		fa[v]=u;S[v].clear();
	}
	for (i=1;i<=n;i++) s+=fa[i]==i;
	puts(s==1?"YES":"NO");
}
```