#include<bits/stdc++.h>
#define For(i,x,y) for (register int i=(x);i<=(y);i++)
#define Dow(i,x,y) for (register int i=(x);i>=(y);i--)
#define cross(i,k) for (register int i=first[k];i;i=last[i])
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<int,int> pa;
inline ll read(){
    ll x=0;int ch=getchar(),f=1;
    while (!isdigit(ch)&&(ch!='-')&&(ch!=EOF)) ch=getchar();
    if (ch=='-'){f=-1;ch=getchar();}
    while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
    return x*f;
}
const int N = 1e5+10;
int n,x,y,h,t;
pa q[N<<1];
map<pa,int>Map;
set<int>s[N];
set<int>::iterator it;
int fa[N];
inline int Find(int x){return fa[x]==x?x:fa[x]=Find(fa[x]);}
inline void Add(int x,int y){s[x].insert(y),s[y].insert(x),Map[mp(x,y)]++,Map[mp(y,x)]++;}
inline void Del(int x,int y){s[x].erase(y),s[y].erase(x),Map[mp(x,y)]=Map[mp(y,x)]=0;}
inline void Merge(int x,int y){
	if (s[y].size()>s[x].size()) swap(x,y);
	while (!s[y].empty()){
		int z=*(s[y].begin());Del(y,z),Add(x,z);
		if (Map[mp(x,z)]==2) q[++t]=mp(x,z);
	}
	fa[y]=x;
}
int main(){
	n=read();
	For(i,1,n-1<<1){
		x=read(),y=read(),Add(x,y);
		if (Map[mp(x,y)]==2) q[++t]=mp(x,y);
	}
	For(i,1,n) fa[i]=i;
	while (h<t){
		int x=Find(q[++h].fi),y=Find(q[h].se);
		s[x].erase(y),s[y].erase(x),Merge(x,y);
	}
	int ans=0;
	For(i,1,n) if (fa[i]==i) ++ans;
	puts(ans==1?"YES":"NO");
}
### Prompt

Please create a solution in CPP 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<bits/stdc++.h>
#define For(i,x,y) for (register int i=(x);i<=(y);i++)
#define Dow(i,x,y) for (register int i=(x);i>=(y);i--)
#define cross(i,k) for (register int i=first[k];i;i=last[i])
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<int,int> pa;
inline ll read(){
    ll x=0;int ch=getchar(),f=1;
    while (!isdigit(ch)&&(ch!='-')&&(ch!=EOF)) ch=getchar();
    if (ch=='-'){f=-1;ch=getchar();}
    while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
    return x*f;
}
const int N = 1e5+10;
int n,x,y,h,t;
pa q[N<<1];
map<pa,int>Map;
set<int>s[N];
set<int>::iterator it;
int fa[N];
inline int Find(int x){return fa[x]==x?x:fa[x]=Find(fa[x]);}
inline void Add(int x,int y){s[x].insert(y),s[y].insert(x),Map[mp(x,y)]++,Map[mp(y,x)]++;}
inline void Del(int x,int y){s[x].erase(y),s[y].erase(x),Map[mp(x,y)]=Map[mp(y,x)]=0;}
inline void Merge(int x,int y){
	if (s[y].size()>s[x].size()) swap(x,y);
	while (!s[y].empty()){
		int z=*(s[y].begin());Del(y,z),Add(x,z);
		if (Map[mp(x,z)]==2) q[++t]=mp(x,z);
	}
	fa[y]=x;
}
int main(){
	n=read();
	For(i,1,n-1<<1){
		x=read(),y=read(),Add(x,y);
		if (Map[mp(x,y)]==2) q[++t]=mp(x,y);
	}
	For(i,1,n) fa[i]=i;
	while (h<t){
		int x=Find(q[++h].fi),y=Find(q[h].se);
		s[x].erase(y),s[y].erase(x),Merge(x,y);
	}
	int ans=0;
	For(i,1,n) if (fa[i]==i) ++ans;
	puts(ans==1?"YES":"NO");
}
```