#include <bits/stdc++.h>
#define N 300010
using namespace std;

inline int read()
{
	int x=0,f=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1; ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}
	return x*f;
}
struct edge
{
	int v,next;
}vs[N<<1];
int n,st[N],ee,fa[N],size[N];
map <int,int> Ms[N];
inline int getfa(int x)
{
	return fa[x]==x? x: fa[x]=getfa(fa[x]);
}
int main()
{
	queue <int> q; 
	while(!q.empty()) q.pop();
	n=read(); for(int i=1;i<n+n-1;i++)
	{
		int u=read(), v=read();
		Ms[u][v]++; Ms[v][u]++;
		size[u]++; size[v]++;
		q.push(u); q.push(v);
	}
	for(int i=1;i<=n;i++) fa[i]=i;
	while(!q.empty())
	{
		int x=q.front(); q.pop();
		int y=q.front(); q.pop();
		x=getfa(x); y=getfa(y);
		if(Ms[x][y]<2) continue;
	//	cerr << x << " " << y << " " << endl;
		if(size[x]<size[y]) swap(x,y);
	//	cout << "merge:" << x << "  " << y << endl;
		for(map<int,int> :: iterator it=Ms[y].begin();it!=Ms[y].end();it++)
		{
			int d=(*it).second,z=(*it).first;
			Ms[y][z]=0; Ms[z][y]=0;
			if(z==x) continue;
			Ms[z][x]+=d; Ms[x][z]+=d;
			q.push(x); q.push(z);
			
		}
		/*for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=n;j++)
				printf("%d ",Ms[i][j]);
			printf("\n");
		}
		cout << endl;*/
		fa[y]=x;
	}
	//for(int i=1;i<=n;i++) cout << fa[i] << " "; cout << endl;
	for(int i=1;i<=n;i++) if(fa[1]!=fa[i]) {puts("NO"); return 0;}
	puts("YES");
	return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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>
#define N 300010
using namespace std;

inline int read()
{
	int x=0,f=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1; ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}
	return x*f;
}
struct edge
{
	int v,next;
}vs[N<<1];
int n,st[N],ee,fa[N],size[N];
map <int,int> Ms[N];
inline int getfa(int x)
{
	return fa[x]==x? x: fa[x]=getfa(fa[x]);
}
int main()
{
	queue <int> q; 
	while(!q.empty()) q.pop();
	n=read(); for(int i=1;i<n+n-1;i++)
	{
		int u=read(), v=read();
		Ms[u][v]++; Ms[v][u]++;
		size[u]++; size[v]++;
		q.push(u); q.push(v);
	}
	for(int i=1;i<=n;i++) fa[i]=i;
	while(!q.empty())
	{
		int x=q.front(); q.pop();
		int y=q.front(); q.pop();
		x=getfa(x); y=getfa(y);
		if(Ms[x][y]<2) continue;
	//	cerr << x << " " << y << " " << endl;
		if(size[x]<size[y]) swap(x,y);
	//	cout << "merge:" << x << "  " << y << endl;
		for(map<int,int> :: iterator it=Ms[y].begin();it!=Ms[y].end();it++)
		{
			int d=(*it).second,z=(*it).first;
			Ms[y][z]=0; Ms[z][y]=0;
			if(z==x) continue;
			Ms[z][x]+=d; Ms[x][z]+=d;
			q.push(x); q.push(z);
			
		}
		/*for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=n;j++)
				printf("%d ",Ms[i][j]);
			printf("\n");
		}
		cout << endl;*/
		fa[y]=x;
	}
	//for(int i=1;i<=n;i++) cout << fa[i] << " "; cout << endl;
	for(int i=1;i<=n;i++) if(fa[1]!=fa[i]) {puts("NO"); return 0;}
	puts("YES");
	return 0;
}
```