#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<set>
#define x first
#define y second

namespace jm
{
	typedef std::pair<int,int> pii;
	const int N=101000;
	std::queue<pii> Q;
	std::set<int> G0[N],G1[N];
	int n;
	void initialize()
	{
		scanf("%d",&n);
		for(int i=1,u,v;i<n;i++)
		{
			scanf("%d%d",&u,&v);
			G0[u].insert(v);
			G0[v].insert(u);
		}

		for(int i=1,u,v;i<n;i++)
		{
			scanf("%d%d",&u,&v);
			G1[u].insert(v);
			G1[v].insert(u);

			if(G0[u].find(v)!=G0[u].end())
				Q.push(pii(u,v));
		}
	}

	bool done[N];

	void combine(int u,int v)
	{
		if(G0[u].size()+G1[u].size()<G0[v].size()+G1[v].size())std::swap(u,v);

		for(auto x:G0[v])if(x!=v)G0[x].erase(v);
		for(auto x:G1[v])if(x!=v)G1[x].erase(v);

		for(auto x:G0[v])
		{
			if(x==u || x==v)continue;

			G0[u].insert(x);
			G0[x].insert(u);
			if(G1[u].count(x))
				Q.push(pii(u,x));
		}

		for(auto x:G1[v])
		{
			if(x==u || x==v)continue;

			G1[u].insert(x);
			G1[x].insert(u);
			if(G0[u].count(x))
				Q.push(pii(u,x));
		}
		G0[v].clear(),G1[v].clear();

		done[v]=1;
	}

	void solve()
	{
		initialize();
//		printf("?\n");
		while(!Q.empty())
		{
			pii A=Q.front();Q.pop();
			int x=A.x,y=A.y;
			if(x==y || done[x] || done[y])continue;

//			printf("%d -- %d\n",x,y);

			combine(x,y),n--;
//			if(n%100==0)printf("%d\n",n);
		}
		printf("%s\n",n>1?"NO":"YES");
	}
}

int main()
{
//	freopen("E.in","r",stdin);
	jm::solve();
	return 0;
}

### Prompt

Generate 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<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<set>
#define x first
#define y second

namespace jm
{
	typedef std::pair<int,int> pii;
	const int N=101000;
	std::queue<pii> Q;
	std::set<int> G0[N],G1[N];
	int n;
	void initialize()
	{
		scanf("%d",&n);
		for(int i=1,u,v;i<n;i++)
		{
			scanf("%d%d",&u,&v);
			G0[u].insert(v);
			G0[v].insert(u);
		}

		for(int i=1,u,v;i<n;i++)
		{
			scanf("%d%d",&u,&v);
			G1[u].insert(v);
			G1[v].insert(u);

			if(G0[u].find(v)!=G0[u].end())
				Q.push(pii(u,v));
		}
	}

	bool done[N];

	void combine(int u,int v)
	{
		if(G0[u].size()+G1[u].size()<G0[v].size()+G1[v].size())std::swap(u,v);

		for(auto x:G0[v])if(x!=v)G0[x].erase(v);
		for(auto x:G1[v])if(x!=v)G1[x].erase(v);

		for(auto x:G0[v])
		{
			if(x==u || x==v)continue;

			G0[u].insert(x);
			G0[x].insert(u);
			if(G1[u].count(x))
				Q.push(pii(u,x));
		}

		for(auto x:G1[v])
		{
			if(x==u || x==v)continue;

			G1[u].insert(x);
			G1[x].insert(u);
			if(G0[u].count(x))
				Q.push(pii(u,x));
		}
		G0[v].clear(),G1[v].clear();

		done[v]=1;
	}

	void solve()
	{
		initialize();
//		printf("?\n");
		while(!Q.empty())
		{
			pii A=Q.front();Q.pop();
			int x=A.x,y=A.y;
			if(x==y || done[x] || done[y])continue;

//			printf("%d -- %d\n",x,y);

			combine(x,y),n--;
//			if(n%100==0)printf("%d\n",n);
		}
		printf("%s\n",n>1?"NO":"YES");
	}
}

int main()
{
//	freopen("E.in","r",stdin);
	jm::solve();
	return 0;
}

```