#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#define LL long long
using namespace std;
vector<int>a,b;
map<LL,int>Hash;
set<int>f[1000005];
int main(void)
{
	int i,j,k,x,y,n,found=0;
	scanf("%d",&n);
	for(i=1;i<=2*n-2;++i)
	{
		scanf("%d%d",&x,&y);
		++Hash[1LL*x*n+y];
		++Hash[1LL*y*n+x];
		f[x].insert(y);
		f[y].insert(x);
		if(Hash[1LL*x*n+y]==2)a.push_back(x),b.push_back(y);
	}
	for(i=0;i<a.size();++i)
	{
		if(Hash[1LL*a[i]*n+b[i]]==2)
		{
			++found;
			if(f[a[i]].size()<f[b[i]].size())swap(a[i],b[i]);
			while(f[b[i]].size())
			{
				j=*f[b[i]].begin();
				Hash[1LL*a[i]*n+j]+=Hash[1LL*b[i]*n+j];
				Hash[1LL*j*n+a[i]]+=Hash[1LL*b[i]*n+j];
				Hash[1LL*b[i]*n+j]=Hash[1LL*j*n+b[i]]=0;
				f[a[i]].insert(j);
				f[j].erase(b[i]);
				f[j].insert(a[i]);
				if(Hash[1LL*a[i]*n+j]==2)a.push_back(a[i]),b.push_back(j);
				f[b[i]].erase(j);
			}
		}
	}
	if(found==n-1)printf("YES\n");
	else printf("NO\n");
	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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#define LL long long
using namespace std;
vector<int>a,b;
map<LL,int>Hash;
set<int>f[1000005];
int main(void)
{
	int i,j,k,x,y,n,found=0;
	scanf("%d",&n);
	for(i=1;i<=2*n-2;++i)
	{
		scanf("%d%d",&x,&y);
		++Hash[1LL*x*n+y];
		++Hash[1LL*y*n+x];
		f[x].insert(y);
		f[y].insert(x);
		if(Hash[1LL*x*n+y]==2)a.push_back(x),b.push_back(y);
	}
	for(i=0;i<a.size();++i)
	{
		if(Hash[1LL*a[i]*n+b[i]]==2)
		{
			++found;
			if(f[a[i]].size()<f[b[i]].size())swap(a[i],b[i]);
			while(f[b[i]].size())
			{
				j=*f[b[i]].begin();
				Hash[1LL*a[i]*n+j]+=Hash[1LL*b[i]*n+j];
				Hash[1LL*j*n+a[i]]+=Hash[1LL*b[i]*n+j];
				Hash[1LL*b[i]*n+j]=Hash[1LL*j*n+b[i]]=0;
				f[a[i]].insert(j);
				f[j].erase(b[i]);
				f[j].insert(a[i]);
				if(Hash[1LL*a[i]*n+j]==2)a.push_back(a[i]),b.push_back(j);
				f[b[i]].erase(j);
			}
		}
	}
	if(found==n-1)printf("YES\n");
	else printf("NO\n");
	return 0;
}

```