#include<cstdio>
#include<map>
#include<set>
#include<iostream>
#define mp(a,b) make_pair(a,b) 
using namespace std;
int fa[100010],n,x,y;
map<pair<int,int>,int>G,vis;
multiset<int> S[100010];
pair<int,int> q[1000010];
int t,h;
void p(int &x,int &y){if (x>y) swap(x,y);} 
int find(int x)
{
	//cout<<x<<endl;
	if (fa[x]!=x) fa[x]=find(fa[x]);
	return fa[x];
}
int main()
{
	scanf("%d",&n);
	for (int i=1;i<n;i++) scanf("%d%d",&x,&y),p(x,y),G[mp(x,y)]++,S[x].insert(y),S[y].insert(x);
	for (int i=1;i<n;i++) {scanf("%d%d",&x,&y),p(x,y);S[x].insert(y),S[y].insert(x);if (++G[mp(x,y)]>1) vis[mp(x,y)]=1,q[++t]=mp(x,y);}
	for (int i=1;i<=n;i++) fa[i]=i;
	for (int k=1;k<n;k++)
	{
		int x,y;
		while (1)
		{
			if (h==t) {printf("NO");return 0;}
			x=find(q[++h].first),y=find(q[h].second);
			if (x==y) continue;
			else break;
		}
		if (S[x].size()>S[y].size()) swap(x,y);
		fa[x]=y;
		for (set<int>::iterator i=S[x].begin();i!=S[x].end();i++)
		{
			int nw=find(*i);
			if (nw==y) continue;
			S[y].insert(nw),S[nw].insert(y);
			int xx,yy;
			if (nw>y) xx=y,yy=nw;
			else xx=nw,yy=y;
			if (++G[mp(xx,yy)]>1&&vis[mp(xx,yy)]!=1) vis[mp(xx,yy)]=1,q[++t]=mp(xx,yy);
			if (S[nw].find(x)!=S[nw].end()) S[nw].erase(x);
		}
		S[y].erase(x); 
	}
	printf("YES");
} 
### 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<map>
#include<set>
#include<iostream>
#define mp(a,b) make_pair(a,b) 
using namespace std;
int fa[100010],n,x,y;
map<pair<int,int>,int>G,vis;
multiset<int> S[100010];
pair<int,int> q[1000010];
int t,h;
void p(int &x,int &y){if (x>y) swap(x,y);} 
int find(int x)
{
	//cout<<x<<endl;
	if (fa[x]!=x) fa[x]=find(fa[x]);
	return fa[x];
}
int main()
{
	scanf("%d",&n);
	for (int i=1;i<n;i++) scanf("%d%d",&x,&y),p(x,y),G[mp(x,y)]++,S[x].insert(y),S[y].insert(x);
	for (int i=1;i<n;i++) {scanf("%d%d",&x,&y),p(x,y);S[x].insert(y),S[y].insert(x);if (++G[mp(x,y)]>1) vis[mp(x,y)]=1,q[++t]=mp(x,y);}
	for (int i=1;i<=n;i++) fa[i]=i;
	for (int k=1;k<n;k++)
	{
		int x,y;
		while (1)
		{
			if (h==t) {printf("NO");return 0;}
			x=find(q[++h].first),y=find(q[h].second);
			if (x==y) continue;
			else break;
		}
		if (S[x].size()>S[y].size()) swap(x,y);
		fa[x]=y;
		for (set<int>::iterator i=S[x].begin();i!=S[x].end();i++)
		{
			int nw=find(*i);
			if (nw==y) continue;
			S[y].insert(nw),S[nw].insert(y);
			int xx,yy;
			if (nw>y) xx=y,yy=nw;
			else xx=nw,yy=y;
			if (++G[mp(xx,yy)]>1&&vis[mp(xx,yy)]!=1) vis[mp(xx,yy)]=1,q[++t]=mp(xx,yy);
			if (S[nw].find(x)!=S[nw].end()) S[nw].erase(x);
		}
		S[y].erase(x); 
	}
	printf("YES");
} 
```