#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<map>
using namespace std;
const int N=100005;
struct node
{
	int x,y;
}t;
bool operator <(node a,node b)
{
	if(a.x!=b.x)
		return a.x<b.x;
	return a.y<b.y;
}
set<node> st;
set<int> e[N];
int n,i,j,k;
map<node,int> mp;
int main()
{
	scanf("%d",&n);
	for(i=1;i<n;++i)
	{
		scanf("%d%d",&j,&k);
		if(j>k)
			swap(j,k);
		++mp[(node){j,k}];
		e[j].insert(k);
		e[k].insert(j);
	}
	for(i=1;i<n;++i)
	{
		scanf("%d%d",&j,&k);
		if(j>k)
			swap(j,k);
		++mp[(node){j,k}];
		e[j].insert(k);
		e[k].insert(j);
	}
	for(map<node,int>::iterator ddq=mp.begin();ddq!=mp.end();ddq++)
		if((ddq->second)>=2)
			st.insert(ddq->first);
	i=0;
	while(!st.empty())
	{
		t=*(st.begin());
		st.erase(t);
		e[t.x].erase(t.y);
		e[t.y].erase(t.x);
		if(e[t.x].size()<e[t.y].size())
			swap(t.x,t.y);
		for(set<int>::iterator ddq=e[t.y].begin();ddq!=e[t.y].end();ddq++)
		{
			e[t.x].insert(*ddq);
			e[*ddq].insert(t.x);
			mp[(node){min(t.x,*ddq),max(t.x,*ddq)}]+=mp[(node){min(t.y,*ddq),max(t.y,*ddq)}];
			if(mp[(node){min(t.y,*ddq),max(t.y,*ddq)}]==2)
				st.erase((node){min(t.y,*ddq),max(t.y,*ddq)});
			if(mp[(node){min(t.x,*ddq),max(t.x,*ddq)}]==2)
				st.insert((node){min(t.x,*ddq),max(t.x,*ddq)});
			e[*ddq].erase(t.y);
		}
		e[t.y].clear();
		++i;
		if(i==n-1)
		{
			printf("YES");
			return 0;
		}
	}
	printf("NO");
	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<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<map>
using namespace std;
const int N=100005;
struct node
{
	int x,y;
}t;
bool operator <(node a,node b)
{
	if(a.x!=b.x)
		return a.x<b.x;
	return a.y<b.y;
}
set<node> st;
set<int> e[N];
int n,i,j,k;
map<node,int> mp;
int main()
{
	scanf("%d",&n);
	for(i=1;i<n;++i)
	{
		scanf("%d%d",&j,&k);
		if(j>k)
			swap(j,k);
		++mp[(node){j,k}];
		e[j].insert(k);
		e[k].insert(j);
	}
	for(i=1;i<n;++i)
	{
		scanf("%d%d",&j,&k);
		if(j>k)
			swap(j,k);
		++mp[(node){j,k}];
		e[j].insert(k);
		e[k].insert(j);
	}
	for(map<node,int>::iterator ddq=mp.begin();ddq!=mp.end();ddq++)
		if((ddq->second)>=2)
			st.insert(ddq->first);
	i=0;
	while(!st.empty())
	{
		t=*(st.begin());
		st.erase(t);
		e[t.x].erase(t.y);
		e[t.y].erase(t.x);
		if(e[t.x].size()<e[t.y].size())
			swap(t.x,t.y);
		for(set<int>::iterator ddq=e[t.y].begin();ddq!=e[t.y].end();ddq++)
		{
			e[t.x].insert(*ddq);
			e[*ddq].insert(t.x);
			mp[(node){min(t.x,*ddq),max(t.x,*ddq)}]+=mp[(node){min(t.y,*ddq),max(t.y,*ddq)}];
			if(mp[(node){min(t.y,*ddq),max(t.y,*ddq)}]==2)
				st.erase((node){min(t.y,*ddq),max(t.y,*ddq)});
			if(mp[(node){min(t.x,*ddq),max(t.x,*ddq)}]==2)
				st.insert((node){min(t.x,*ddq),max(t.x,*ddq)});
			e[*ddq].erase(t.y);
		}
		e[t.y].clear();
		++i;
		if(i==n-1)
		{
			printf("YES");
			return 0;
		}
	}
	printf("NO");
	return 0;
}

```