#include<iostream>
#include<vector>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<set>
using namespace std;
typedef long long LL;
const int N=2e5;
int gi() {
	int w=0;bool q=1;char c=getchar();
	while ((c<'0'||c>'9') && c!='-') c=getchar();
	if (c=='-') q=0,c=getchar();
	while (c>='0'&&c <= '9') w=w*10+c-'0',c=getchar();
	return q? w:-w;
}
int fa[N],siz[N];
inline int find(int x) { return fa[x]==x?x:fa[x]=find(fa[x]); }
multiset<int>e[N];
int main()
{
	int n=gi(),i,a,b,t=n-1,k;pair<int,int>p;
	multiset< pair<int,int> >s;multiset<int>::iterator it;
	queue< pair<int,int> >q;
	for (i=(n-1)<<1;i;i--) {
		if ((a=gi())>(b=gi())) swap(a,b);
		e[a].insert(b);
		e[b].insert(a);
		s.insert(make_pair(a,b));
		if (s.count(make_pair(a,b))==2) q.push(make_pair(a,b)),s.erase(make_pair(a,b));
	}
	for (i=1;i<=n;i++) fa[i]=i,siz[i]=1;
	while (!q.empty()) {
		a=q.front().first;
		b=q.front().second;
		q.pop();
		if (fa[a]!=a||fa[b]!=b||a==b) continue;t--;
		if (siz[fa[a]]>siz[fa[b]]) swap(a,b);
		siz[b]+=siz[a];fa[a]=b;
		e[b].erase(a);
		e[a].erase(b);
		for (it=e[a].begin();it!=e[a].end();it++) {
			e[*it].erase(e[*it].find(a));
			k=find(*it);
			e[k].insert(b);
			e[b].insert(k);
			p=make_pair(min(k,b),max(k,b));
			s.insert(p);
			if (s.count(p)==2) s.erase(p),q.push(p);
		}
	}
	puts(t?"NO":"YES");
	return 0;
}

### Prompt

Your task is to create 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<vector>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<set>
using namespace std;
typedef long long LL;
const int N=2e5;
int gi() {
	int w=0;bool q=1;char c=getchar();
	while ((c<'0'||c>'9') && c!='-') c=getchar();
	if (c=='-') q=0,c=getchar();
	while (c>='0'&&c <= '9') w=w*10+c-'0',c=getchar();
	return q? w:-w;
}
int fa[N],siz[N];
inline int find(int x) { return fa[x]==x?x:fa[x]=find(fa[x]); }
multiset<int>e[N];
int main()
{
	int n=gi(),i,a,b,t=n-1,k;pair<int,int>p;
	multiset< pair<int,int> >s;multiset<int>::iterator it;
	queue< pair<int,int> >q;
	for (i=(n-1)<<1;i;i--) {
		if ((a=gi())>(b=gi())) swap(a,b);
		e[a].insert(b);
		e[b].insert(a);
		s.insert(make_pair(a,b));
		if (s.count(make_pair(a,b))==2) q.push(make_pair(a,b)),s.erase(make_pair(a,b));
	}
	for (i=1;i<=n;i++) fa[i]=i,siz[i]=1;
	while (!q.empty()) {
		a=q.front().first;
		b=q.front().second;
		q.pop();
		if (fa[a]!=a||fa[b]!=b||a==b) continue;t--;
		if (siz[fa[a]]>siz[fa[b]]) swap(a,b);
		siz[b]+=siz[a];fa[a]=b;
		e[b].erase(a);
		e[a].erase(b);
		for (it=e[a].begin();it!=e[a].end();it++) {
			e[*it].erase(e[*it].find(a));
			k=find(*it);
			e[k].insert(b);
			e[b].insert(k);
			p=make_pair(min(k,b),max(k,b));
			s.insert(p);
			if (s.count(p)==2) s.erase(p),q.push(p);
		}
	}
	puts(t?"NO":"YES");
	return 0;
}

```