#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<set>
#include<map>
#include<queue>
#define SF scanf
#define PF printf
#define MAXN 100010
using namespace std;
typedef long long ll;
set<int> st[MAXN];
map<ll,int> mp;
int n,fa[MAXN];
ll get_num(int x,int y){
	if(x>y)
		swap(x,y);
	return 1ll*(x-1)*n+y-1;
}
queue<ll> q;
void add_edge(int x,int y){
	ll id=get_num(x,y);
	mp[id]++;
	if(mp[id]==2)
		q.push(id);
	if(mp[id]>2){
		PF("NO");
		exit(0);	
	}
	st[x].insert(y);
	st[y].insert(x);
}
void Del(int x,int y){
	set<int>::iterator it=st[x].begin();
	for(;it!=st[x].end();it++){
		int u=*it;
		st[u].erase(x);
		if(u!=y){
			st[u].insert(y);
			st[y].insert(u);
			mp[get_num(u,y)]++;
			if(mp[get_num(u,y)]==2)
				q.push(get_num(u,y));
			if(mp[get_num(u,y)]>2){
				PF("NO!");
				exit(0);	
			}
		}
		mp.erase(get_num(u,x));	
	}
	st[x].clear();
}
int get_fa(int x){
	if(fa[x]==0)
		return x;
	fa[x]=get_fa(fa[x]);
	return fa[x];	
}
int main(){
	//freopen("bluered.in","r",stdin);
	//freopen("bluered.out","w",stdout);
	SF("%d",&n);
	int u,v;
	for(int i=1;i<2*n-1;i++){
		SF("%d%d",&u,&v);
		add_edge(u,v);	
	}
	for(int i=1;i<n;i++){
		if(q.empty()){
			PF("NO");
			return 0;	
		}
		ll id=q.front();
		q.pop();
		int u=id%n+1;
		int v=id/n+1;
		if(u==v){
			i--;
			continue;
		}
		u=get_fa(u);
		v=get_fa(v);
		if(st[u].size()>st[v].size())
			swap(u,v);
		Del(u,v);
		fa[u]=v;
	}
	PF("YES");
}
### Prompt

In cpp, your task is to solve 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<cmath>
#include<set>
#include<map>
#include<queue>
#define SF scanf
#define PF printf
#define MAXN 100010
using namespace std;
typedef long long ll;
set<int> st[MAXN];
map<ll,int> mp;
int n,fa[MAXN];
ll get_num(int x,int y){
	if(x>y)
		swap(x,y);
	return 1ll*(x-1)*n+y-1;
}
queue<ll> q;
void add_edge(int x,int y){
	ll id=get_num(x,y);
	mp[id]++;
	if(mp[id]==2)
		q.push(id);
	if(mp[id]>2){
		PF("NO");
		exit(0);	
	}
	st[x].insert(y);
	st[y].insert(x);
}
void Del(int x,int y){
	set<int>::iterator it=st[x].begin();
	for(;it!=st[x].end();it++){
		int u=*it;
		st[u].erase(x);
		if(u!=y){
			st[u].insert(y);
			st[y].insert(u);
			mp[get_num(u,y)]++;
			if(mp[get_num(u,y)]==2)
				q.push(get_num(u,y));
			if(mp[get_num(u,y)]>2){
				PF("NO!");
				exit(0);	
			}
		}
		mp.erase(get_num(u,x));	
	}
	st[x].clear();
}
int get_fa(int x){
	if(fa[x]==0)
		return x;
	fa[x]=get_fa(fa[x]);
	return fa[x];	
}
int main(){
	//freopen("bluered.in","r",stdin);
	//freopen("bluered.out","w",stdout);
	SF("%d",&n);
	int u,v;
	for(int i=1;i<2*n-1;i++){
		SF("%d%d",&u,&v);
		add_edge(u,v);	
	}
	for(int i=1;i<n;i++){
		if(q.empty()){
			PF("NO");
			return 0;	
		}
		ll id=q.front();
		q.pop();
		int u=id%n+1;
		int v=id/n+1;
		if(u==v){
			i--;
			continue;
		}
		u=get_fa(u);
		v=get_fa(v);
		if(st[u].size()>st[v].size())
			swap(u,v);
		Del(u,v);
		fa[u]=v;
	}
	PF("YES");
}
```