#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define MAX 100100
#define pi pair<int,int>
#define mp make_pair
#define fr first
#define sd second
inline int read()
{
	int x=0;bool t=false;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')t=true,ch=getchar();
	while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
	return t?-x:x;
}
int n,f[MAX];
int getf(int x){return x==f[x]?x:f[x]=getf(f[x]);}
queue<pi> Q;
map<pi,int> M;
set<int> E[MAX];
pi get(int x,int y){if(x>y)swap(x,y);return mp(x,y);}
void Link(int x,int y)
{
	E[x].insert(y);E[y].insert(x);
	pi u=get(x,y);
	if(++M[u]==2)Q.push(u);
}
int main()
{
	n=read();
	for(int i=1;i<=n;++i)f[i]=i;
	for(int i=1;i<=n+n-2;++i)Link(read(),read());
	for(int i=1,x,y;i<n;++i)
	{
		do
		{
			if(Q.empty()){puts("NO");return 0;}
			x=Q.front().fr,y=Q.front().sd;Q.pop();
			x=getf(x);y=getf(y);
		}while(x==y);
		if(E[x].size()>E[y].size())swap(x,y);
		f[x]=y;M.erase(get(x,y));E[y].erase(x);
		for(auto v:E[x])
		{
			int fv=getf(v);
			if(fv==y)continue;
			M.erase(get(x,fv));
			E[fv].erase(x);//E[x].erase(fv);
			Link(fv,y);
		}
	}
	puts("YES");
	return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define MAX 100100
#define pi pair<int,int>
#define mp make_pair
#define fr first
#define sd second
inline int read()
{
	int x=0;bool t=false;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')t=true,ch=getchar();
	while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
	return t?-x:x;
}
int n,f[MAX];
int getf(int x){return x==f[x]?x:f[x]=getf(f[x]);}
queue<pi> Q;
map<pi,int> M;
set<int> E[MAX];
pi get(int x,int y){if(x>y)swap(x,y);return mp(x,y);}
void Link(int x,int y)
{
	E[x].insert(y);E[y].insert(x);
	pi u=get(x,y);
	if(++M[u]==2)Q.push(u);
}
int main()
{
	n=read();
	for(int i=1;i<=n;++i)f[i]=i;
	for(int i=1;i<=n+n-2;++i)Link(read(),read());
	for(int i=1,x,y;i<n;++i)
	{
		do
		{
			if(Q.empty()){puts("NO");return 0;}
			x=Q.front().fr,y=Q.front().sd;Q.pop();
			x=getf(x);y=getf(y);
		}while(x==y);
		if(E[x].size()>E[y].size())swap(x,y);
		f[x]=y;M.erase(get(x,y));E[y].erase(x);
		for(auto v:E[x])
		{
			int fv=getf(v);
			if(fv==y)continue;
			M.erase(get(x,fv));
			E[fv].erase(x);//E[x].erase(fv);
			Link(fv,y);
		}
	}
	puts("YES");
	return 0;
}

```