#include<bits/stdc++.h>
#define reg register
typedef long long ll;
using namespace std;
const int MN=1e5+5;
typedef pair<int,int>pii;
#define fi first
#define se second
set<int>node[MN];
map<pii,int>mp;
queue<pii>q;
int n,par[MN];
inline int find(int x){return par[x]==x?x:par[x]=find(par[x]);}
inline void del(int x,int y){
	node[x].erase(y);mp[make_pair(min(x,y),max(x,y))]=0;
}
inline void add(int x,int y){
	if(x>y)swap(x,y);node[x].insert(y);node[y].insert(x);
	if((++mp[make_pair(x,y)])==2)q.push(make_pair(x,y));
}
int main(){
	scanf("%d",&n);
	for(reg int i=1,s,t;i<n+n-1;i++)
		scanf("%d%d",&s,&t),add(s,t);
	for(reg int i=1;i<=n;i++)par[i]=i;
	while(!q.empty()){
		reg pii v=q.front();q.pop();
		reg int x=find(v.fi),y=find(v.se);
		node[x].erase(y);node[y].erase(x);
		if(node[x].size()<node[y].size())swap(x,y);
		for(auto t:node[y])del(t,y),add(t,x);
		par[y]=x;node[y].clear();
	}
	for(reg int i=2;i<=n;i++)
		if(find(i)!=find(1))return 0*puts("NO");
	return 0*puts("YES");
}
### 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>
#define reg register
typedef long long ll;
using namespace std;
const int MN=1e5+5;
typedef pair<int,int>pii;
#define fi first
#define se second
set<int>node[MN];
map<pii,int>mp;
queue<pii>q;
int n,par[MN];
inline int find(int x){return par[x]==x?x:par[x]=find(par[x]);}
inline void del(int x,int y){
	node[x].erase(y);mp[make_pair(min(x,y),max(x,y))]=0;
}
inline void add(int x,int y){
	if(x>y)swap(x,y);node[x].insert(y);node[y].insert(x);
	if((++mp[make_pair(x,y)])==2)q.push(make_pair(x,y));
}
int main(){
	scanf("%d",&n);
	for(reg int i=1,s,t;i<n+n-1;i++)
		scanf("%d%d",&s,&t),add(s,t);
	for(reg int i=1;i<=n;i++)par[i]=i;
	while(!q.empty()){
		reg pii v=q.front();q.pop();
		reg int x=find(v.fi),y=find(v.se);
		node[x].erase(y);node[y].erase(x);
		if(node[x].size()<node[y].size())swap(x,y);
		for(auto t:node[y])del(t,y),add(t,x);
		par[y]=x;node[y].clear();
	}
	for(reg int i=2;i<=n;i++)
		if(find(i)!=find(1))return 0*puts("NO");
	return 0*puts("YES");
}
```