#include<cstdio>
#include<cstring>
#include<algorithm>
#include<set>
#define rep(a,b,c) for (int a=b;a<=c;a++)
#define per(a,b,c) for (int a=b;a>=c;a--)
#define go(u) for (int o=ft[u],v;v=E[o].t;o=E[o].n)
#define fi first
#define se second
using namespace std;
typedef long long LL;
typedef pair<int,int> par;
const int N=100010;
set <int> s[N],s2[N];
set <int> :: iterator it;
int n,t,fa[N];
par q[N<<4];
int get(int x){return fa[x]?fa[x]=get(fa[x]):x;}
int merge(int u,int v){
	if ((u=get(u))==(v=get(v))) return 0;
	if (s[u].size()+s2[u].size()<s[v].size()+s2[v].size()) swap(u,v);
	fa[v]=u;
	s[u].erase(s[u].find(v));
	s2[u].erase(s2[u].find(v));
	for (it=s[v].begin();it!=s[v].end();++it){
		if (*it==u) continue;
		int x=*it;
		s[u].insert(x);
		s[x].erase(s[x].find(v));
		s[x].insert(u);
		if (s2[x].find(u)!=s2[x].end()) q[++t]=(par){x,u};
	}
	for (it=s2[v].begin();it!=s2[v].end();++it){
		if (*it==u) continue;
		int x=*it;
		s2[u].insert(x);
		s2[x].erase(s2[x].find(v));
		s2[x].insert(u);
		if (s[x].find(u)!=s[x].end()) q[++t]=(par){x,u};
	}
	return 1;
}
int main(){
	scanf("%d",&n);
	rep(i,2,n){
		int x,y;
		scanf("%d%d",&x,&y);
		s[x].insert(y);
		s[y].insert(x);
	}
	rep(i,2,n){
		int x,y;
		scanf("%d%d",&x,&y);
		s2[x].insert(y);
		s2[y].insert(x);
		if (s[x].find(y)!=s[x].end()) q[++t]=(par){x,y};
	}
	int tot=n;
	rep(h,1,t) tot-=merge(q[h].fi,q[h].se);
	if (tot==1) puts("YES"); else puts("NO");
	return 0;
}

### Prompt

Please create a solution in CPP 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<cstdio>
#include<cstring>
#include<algorithm>
#include<set>
#define rep(a,b,c) for (int a=b;a<=c;a++)
#define per(a,b,c) for (int a=b;a>=c;a--)
#define go(u) for (int o=ft[u],v;v=E[o].t;o=E[o].n)
#define fi first
#define se second
using namespace std;
typedef long long LL;
typedef pair<int,int> par;
const int N=100010;
set <int> s[N],s2[N];
set <int> :: iterator it;
int n,t,fa[N];
par q[N<<4];
int get(int x){return fa[x]?fa[x]=get(fa[x]):x;}
int merge(int u,int v){
	if ((u=get(u))==(v=get(v))) return 0;
	if (s[u].size()+s2[u].size()<s[v].size()+s2[v].size()) swap(u,v);
	fa[v]=u;
	s[u].erase(s[u].find(v));
	s2[u].erase(s2[u].find(v));
	for (it=s[v].begin();it!=s[v].end();++it){
		if (*it==u) continue;
		int x=*it;
		s[u].insert(x);
		s[x].erase(s[x].find(v));
		s[x].insert(u);
		if (s2[x].find(u)!=s2[x].end()) q[++t]=(par){x,u};
	}
	for (it=s2[v].begin();it!=s2[v].end();++it){
		if (*it==u) continue;
		int x=*it;
		s2[u].insert(x);
		s2[x].erase(s2[x].find(v));
		s2[x].insert(u);
		if (s[x].find(u)!=s[x].end()) q[++t]=(par){x,u};
	}
	return 1;
}
int main(){
	scanf("%d",&n);
	rep(i,2,n){
		int x,y;
		scanf("%d%d",&x,&y);
		s[x].insert(y);
		s[y].insert(x);
	}
	rep(i,2,n){
		int x,y;
		scanf("%d%d",&x,&y);
		s2[x].insert(y);
		s2[y].insert(x);
		if (s[x].find(y)!=s[x].end()) q[++t]=(par){x,y};
	}
	int tot=n;
	rep(h,1,t) tot-=merge(q[h].fi,q[h].se);
	if (tot==1) puts("YES"); else puts("NO");
	return 0;
}

```