#include<bits/stdc++.h> 
#define fo(i,a,b) for(i=a;i<=b;i++)
#define fd(i,a,b) for(i=a;i>=b;i--)
#define min(a,b) (a<b?a:b)
#define max(a,b) (a>b?a:b)
typedef long long ll;
using namespace std;
inline int read(){
	int n=0,f=1;char c;
	for(c=getchar();(c<'0'||c>'9')&&c!='-';c=getchar());
	if (c=='-') c=getchar(),f=-1;
	for(;c>='0'&&c<='9';c=getchar()) n=n*10+c-48;
	return n*f;
}
const int maxn=1e5+5;
multiset<int>p[maxn];
struct ar{int x,y;}now;
queue<ar>a;
map<ll,int>w;
int i,j,n,f[maxn];
int get(int x){
	return f[x]?f[x]=get(f[x]):x;
}
ll pd(int x,int y){
	if (x>y) swap(x,y);
	return (ll)x*maxn+y;
}
void add(int x,int y){
	p[x].insert(y);
	p[y].insert(x);
	ll s=pd(x,y);
	w[s]++;
	if (w[s]==2) a.push((ar){x,y});
	if (w[s]==3) puts("NO"),exit(0); 
}
int main(){
	n=read();
	fo(i,1,(n-1)<<1){
		int x=read(),y=read();
		add(x,y);
	}
	n--;
	for(;!a.empty();){
		now=a.front(),a.pop();
		int x=get(now.x),y=get(now.y);
		if (x!=now.x||y!=now.y) continue;
		n--;
		if (p[x].size()>p[y].size()) swap(x,y);
		f[x]=y,p[y].erase(x),p[x].erase(y);		
		set<int>::iterator it;
		for(it=p[x].begin();it!=p[x].end();it++){
			int z=get(*it);
			add(y,z),p[z].erase(x);
		}
	}
	if (n) puts("NO");else puts("YES");
}
### 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 fo(i,a,b) for(i=a;i<=b;i++)
#define fd(i,a,b) for(i=a;i>=b;i--)
#define min(a,b) (a<b?a:b)
#define max(a,b) (a>b?a:b)
typedef long long ll;
using namespace std;
inline int read(){
	int n=0,f=1;char c;
	for(c=getchar();(c<'0'||c>'9')&&c!='-';c=getchar());
	if (c=='-') c=getchar(),f=-1;
	for(;c>='0'&&c<='9';c=getchar()) n=n*10+c-48;
	return n*f;
}
const int maxn=1e5+5;
multiset<int>p[maxn];
struct ar{int x,y;}now;
queue<ar>a;
map<ll,int>w;
int i,j,n,f[maxn];
int get(int x){
	return f[x]?f[x]=get(f[x]):x;
}
ll pd(int x,int y){
	if (x>y) swap(x,y);
	return (ll)x*maxn+y;
}
void add(int x,int y){
	p[x].insert(y);
	p[y].insert(x);
	ll s=pd(x,y);
	w[s]++;
	if (w[s]==2) a.push((ar){x,y});
	if (w[s]==3) puts("NO"),exit(0); 
}
int main(){
	n=read();
	fo(i,1,(n-1)<<1){
		int x=read(),y=read();
		add(x,y);
	}
	n--;
	for(;!a.empty();){
		now=a.front(),a.pop();
		int x=get(now.x),y=get(now.y);
		if (x!=now.x||y!=now.y) continue;
		n--;
		if (p[x].size()>p[y].size()) swap(x,y);
		f[x]=y,p[y].erase(x),p[x].erase(y);		
		set<int>::iterator it;
		for(it=p[x].begin();it!=p[x].end();it++){
			int z=get(*it);
			add(y,z),p[z].erase(x);
		}
	}
	if (n) puts("NO");else puts("YES");
}
```