// luogu-judger-enable-o2
#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<ctime>
#include<set>
#include<map>
#include<climits>
#include<algorithm>
#include<queue>
#define ll long long
using namespace std;
const int N=100010;
typedef set<int>::iterator It;
set<int> s[N];
map<ll,int> ma;
queue<pair<int,int> > q;
int n,f[N];
int find(int x) {return f[x]==x?x:f[x]=find(f[x]);}
ll change(int x,int y) {return (ll)min(x,y)*n+max(x,y);}
void erase(int x,int y)
{
	s[x].erase(y);
	s[y].erase(x);
	ma.erase(change(x,y));
}
void insert(int x,int y)
{
	s[x].insert(y);
	s[y].insert(x);
	ma[change(x,y)]++;
	if(ma[change(x,y)]==2) q.push(make_pair(x,y));
}
int main()
{
	int i,j,x,y,t1,t2;
	ll t;
	scanf("%d",&n);
	for(i=1;i<=n;i++) f[i]=i;
	for(i=1;i<2*n-1;i++)
	{
	    scanf("%d%d",&x,&y);
		insert(x,y);
	}
	for(i=1;i<n;i++)
	{
		if(q.empty()) break;
		x=q.front().first,y=q.front().second,q.pop();
		t1=find(x),t2=find(y);
		if(s[t1].size()<s[t2].size()) swap(t1,t2),swap(x,y);
		erase(t1,t2);
		f[t2]=t1;
		It it,j;
		for(It it=s[t2].begin();it!=s[t2].end();it=j) 
		{
			j=it;j++;
			erase(t2,find(*it));
			insert(t1,find(*it));
		}
	}
	if(i==n) printf("YES");
	else printf("NO");
	return 0;
}
### Prompt

Your challenge is to write 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
// luogu-judger-enable-o2
#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<ctime>
#include<set>
#include<map>
#include<climits>
#include<algorithm>
#include<queue>
#define ll long long
using namespace std;
const int N=100010;
typedef set<int>::iterator It;
set<int> s[N];
map<ll,int> ma;
queue<pair<int,int> > q;
int n,f[N];
int find(int x) {return f[x]==x?x:f[x]=find(f[x]);}
ll change(int x,int y) {return (ll)min(x,y)*n+max(x,y);}
void erase(int x,int y)
{
	s[x].erase(y);
	s[y].erase(x);
	ma.erase(change(x,y));
}
void insert(int x,int y)
{
	s[x].insert(y);
	s[y].insert(x);
	ma[change(x,y)]++;
	if(ma[change(x,y)]==2) q.push(make_pair(x,y));
}
int main()
{
	int i,j,x,y,t1,t2;
	ll t;
	scanf("%d",&n);
	for(i=1;i<=n;i++) f[i]=i;
	for(i=1;i<2*n-1;i++)
	{
	    scanf("%d%d",&x,&y);
		insert(x,y);
	}
	for(i=1;i<n;i++)
	{
		if(q.empty()) break;
		x=q.front().first,y=q.front().second,q.pop();
		t1=find(x),t2=find(y);
		if(s[t1].size()<s[t2].size()) swap(t1,t2),swap(x,y);
		erase(t1,t2);
		f[t2]=t1;
		It it,j;
		for(It it=s[t2].begin();it!=s[t2].end();it=j) 
		{
			j=it;j++;
			erase(t2,find(*it));
			insert(t1,find(*it));
		}
	}
	if(i==n) printf("YES");
	else printf("NO");
	return 0;
}
```