

#include<bits/stdc++.h>
const int MN=100000+5;
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
template<typename T>T&IN(T&in){
	in=0;char c=getchar();int f=1;
	while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
	while(isdigit(c))in=in*10+c-'0',c=getchar();
	return in*=f;
}
int n;
int fa[MN];
set<int>h[MN];
set<int>::iterator it;
map<pii,int>mp;
queue<pii>q;pii e;
int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}
void add(int u,int v){
	if(u>v)swap(u,v);
	h[u].insert(v),h[v].insert(u);mp[pii(u,v)]++;
	if(mp[pii(u,v)]==2)q.push(pii(u,v));
}
void input(){
	IN(n);
	for(int i=1;i<=n;++i)fa[i]=i;
	for(int i=1,u,v;i<=2*n-2;++i)IN(u),IN(v),add(u,v);
	for(int i=1,u,v;i<n;++i){
		while(1){
			if(q.empty())return puts("NO"),void();
			e=q.front();q.pop();u=find(e.first),v=find(e.second);
			if(u!=v)break;
		}
		if(h[u].size()>h[v].size())swap(u,v);
		fa[u]=v;mp.erase(e),h[v].erase(u);
		for(it=h[u].begin();it!=h[u].end();it++){
			int x=find(*it);if(x==v)continue;
			add(x,v);
		}
	}
	puts("YES");
}
int main(){
	input();
	return 0;
}

### Prompt

Create a solution in CPP for 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>
const int MN=100000+5;
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
template<typename T>T&IN(T&in){
	in=0;char c=getchar();int f=1;
	while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
	while(isdigit(c))in=in*10+c-'0',c=getchar();
	return in*=f;
}
int n;
int fa[MN];
set<int>h[MN];
set<int>::iterator it;
map<pii,int>mp;
queue<pii>q;pii e;
int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}
void add(int u,int v){
	if(u>v)swap(u,v);
	h[u].insert(v),h[v].insert(u);mp[pii(u,v)]++;
	if(mp[pii(u,v)]==2)q.push(pii(u,v));
}
void input(){
	IN(n);
	for(int i=1;i<=n;++i)fa[i]=i;
	for(int i=1,u,v;i<=2*n-2;++i)IN(u),IN(v),add(u,v);
	for(int i=1,u,v;i<n;++i){
		while(1){
			if(q.empty())return puts("NO"),void();
			e=q.front();q.pop();u=find(e.first),v=find(e.second);
			if(u!=v)break;
		}
		if(h[u].size()>h[v].size())swap(u,v);
		fa[u]=v;mp.erase(e),h[v].erase(u);
		for(it=h[u].begin();it!=h[u].end();it++){
			int x=find(*it);if(x==v)continue;
			add(x,v);
		}
	}
	puts("YES");
}
int main(){
	input();
	return 0;
}

```