#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<stdlib.h>
#include<string.h>
#include<queue>
#include<map>
#include<set>
#define LL long long
#define fr first
#define sc second
#define SX E[x].size( )
#define SY E[y].size( )
using namespace std;
typedef pair<int,int> pr;
inline pr mp(int x,int y) {return x<y?make_pair(x,y):make_pair(y,x);}
inline int read( )
{
  int sum=0;char c=getchar( );bool f=0;
  while(c<'0'||c>'9') {if(c=='-') f=1;c=getchar( );}
  while(c>='0'&&c<='9') {sum=sum*10+c-'0';c=getchar( );}
  if(f) return -sum;
  return sum;
}
const int N=100005;
int n,m,f[N];
map<pr,int>H;set<int>E[N];queue<pr>q;
inline int find(int x) {return f[x]==x?x:f[x]=find(f[x]);}
int main( )
{
  int i,x,y,k;pr o;
  n=read( );m=2*n-2;
  for(i=1;i<=m;i++)
    {
      x=read( );y=read( );
      E[x].insert(y);E[y].insert(x);
      if((++H[o=mp(x,y)])==2) q.push(o);
    }
  for(i=1;i<=n;i++) f[i]=i;
  set<int>::iterator it,ed;
  while(!q.empty( ))
    {
      o=q.front( );q.pop( );
      x=find(o.fr);y=find(o.sc);
      if(x==y) continue;n--;
      if(SX>SY) swap(x,y);
      for(it=E[x].begin( ),ed=E[x].end( );it!=ed;it++)
	{
	  k=find(*it);
	  if(k==y) continue;
	  else E[y].insert(k);
	  --H[mp(k,x)];
	  if((++H[o=mp(k,y)])==2) q.push(o);
	}
      f[x]=y;
    }
  puts(n==1?"YES":"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<stdio.h>
#include<iostream>
#include<algorithm>
#include<stdlib.h>
#include<string.h>
#include<queue>
#include<map>
#include<set>
#define LL long long
#define fr first
#define sc second
#define SX E[x].size( )
#define SY E[y].size( )
using namespace std;
typedef pair<int,int> pr;
inline pr mp(int x,int y) {return x<y?make_pair(x,y):make_pair(y,x);}
inline int read( )
{
  int sum=0;char c=getchar( );bool f=0;
  while(c<'0'||c>'9') {if(c=='-') f=1;c=getchar( );}
  while(c>='0'&&c<='9') {sum=sum*10+c-'0';c=getchar( );}
  if(f) return -sum;
  return sum;
}
const int N=100005;
int n,m,f[N];
map<pr,int>H;set<int>E[N];queue<pr>q;
inline int find(int x) {return f[x]==x?x:f[x]=find(f[x]);}
int main( )
{
  int i,x,y,k;pr o;
  n=read( );m=2*n-2;
  for(i=1;i<=m;i++)
    {
      x=read( );y=read( );
      E[x].insert(y);E[y].insert(x);
      if((++H[o=mp(x,y)])==2) q.push(o);
    }
  for(i=1;i<=n;i++) f[i]=i;
  set<int>::iterator it,ed;
  while(!q.empty( ))
    {
      o=q.front( );q.pop( );
      x=find(o.fr);y=find(o.sc);
      if(x==y) continue;n--;
      if(SX>SY) swap(x,y);
      for(it=E[x].begin( ),ed=E[x].end( );it!=ed;it++)
	{
	  k=find(*it);
	  if(k==y) continue;
	  else E[y].insert(k);
	  --H[mp(k,x)];
	  if((++H[o=mp(k,y)])==2) q.push(o);
	}
      f[x]=y;
    }
  puts(n==1?"YES":"NO");
  return 0;
}

```