#include <set>
#include <queue>
#include <cstdio>
#include <algorithm>

using namespace std;

const int N=1e5+10;

int n,f[N];
set<int>B[N],R[N];
queue< pair<int,int> >Q;

void Init()
{
	scanf("%d",&n);
	int a,b;
	for (int i=1;i<n;++i)
	{
		scanf("%d%d",&a,&b);
		B[a].insert(b); B[b].insert(a);
	}
	for (int i=1;i<n;++i)
	{
		scanf("%d%d",&a,&b);
		R[a].insert(b); R[b].insert(a); 
		if (B[a].count(b)) Q.push(make_pair(a,b));
	}
}

int find(int a) { return a==f[a]?a:f[a]=find(f[a]); }

void Union(int x,int y)
{
	if (R[x].size()>R[y].size()) swap(x,y);
	f[x]=y;
	for (auto a:R[x]) if (B[a].count(y)) Q.push(make_pair(a,y));
	for (auto a:B[x]) if (R[a].count(y)) Q.push(make_pair(a,y));
	for (auto a:R[x]) { R[a].erase(x); R[a].insert(y); R[y].insert(a); }
	for (auto a:B[x]) { B[a].erase(x); B[a].insert(y); B[y].insert(a); }
}

void Solve()
{
	for (int i=1;i<=n;++i) f[i]=i;
	int cnt=0;
	while (!Q.empty())
	{
		int x=Q.front().first,y=Q.front().second; Q.pop();
		x=find(x); y=find(y);
		if (x==y)  continue;
		++cnt;
		R[x].erase(y); R[y].erase(x); B[x].erase(y); B[y].erase(x);
		Union(x,y);
	}
	if (cnt==n-1) printf("YES\n");
	else printf("NO\n");
}

int main()
{
	Init();
	Solve();
	return 0;
}
### Prompt

Develop a solution in Cpp 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 <set>
#include <queue>
#include <cstdio>
#include <algorithm>

using namespace std;

const int N=1e5+10;

int n,f[N];
set<int>B[N],R[N];
queue< pair<int,int> >Q;

void Init()
{
	scanf("%d",&n);
	int a,b;
	for (int i=1;i<n;++i)
	{
		scanf("%d%d",&a,&b);
		B[a].insert(b); B[b].insert(a);
	}
	for (int i=1;i<n;++i)
	{
		scanf("%d%d",&a,&b);
		R[a].insert(b); R[b].insert(a); 
		if (B[a].count(b)) Q.push(make_pair(a,b));
	}
}

int find(int a) { return a==f[a]?a:f[a]=find(f[a]); }

void Union(int x,int y)
{
	if (R[x].size()>R[y].size()) swap(x,y);
	f[x]=y;
	for (auto a:R[x]) if (B[a].count(y)) Q.push(make_pair(a,y));
	for (auto a:B[x]) if (R[a].count(y)) Q.push(make_pair(a,y));
	for (auto a:R[x]) { R[a].erase(x); R[a].insert(y); R[y].insert(a); }
	for (auto a:B[x]) { B[a].erase(x); B[a].insert(y); B[y].insert(a); }
}

void Solve()
{
	for (int i=1;i<=n;++i) f[i]=i;
	int cnt=0;
	while (!Q.empty())
	{
		int x=Q.front().first,y=Q.front().second; Q.pop();
		x=find(x); y=find(y);
		if (x==y)  continue;
		++cnt;
		R[x].erase(y); R[y].erase(x); B[x].erase(y); B[y].erase(x);
		Union(x,y);
	}
	if (cnt==n-1) printf("YES\n");
	else printf("NO\n");
}

int main()
{
	Init();
	Solve();
	return 0;
}
```