#pragma GCC optimize(2)
#include <cstdio>
#include <vector>
#include <map>
#include <set>
#define N 1000011
using namespace std;
struct point{
	int x,y;
	point(){}
	point(int xx,int yy){x=xx;y=yy;}
	bool operator<(const point &other)const{
		if(x==other.x) return y<other.y;
		return x<other.x;
	}
}que[N];
set<int> V[N];
set<int>::iterator it;
map<point,int> Map;
int father[N],head,tail,n,ans;
inline int read(){
	int ret=0;char ch=getchar();
	while(ch<'0' || ch>'9') ch=getchar();
	while(ch>='0'&&ch<='9'){ret=ret*10+ch-'0';ch=getchar();}
	return ret;
}
int find(int u){
	if(u==father[u]) return u;
	father[u]=find(father[u]);
	return father[u];
}
int main(){
	int x,y;
	scanf("%d",&n);
	for(int i=1;i<=n;i++) father[i]=i;
	for(int i=1;i<=n*2-2;i++){
		x=read(),y=read();
		Map[point(x,y)]++;
		Map[point(y,x)]++;
		V[x].insert(y);
		V[y].insert(x);
		if(Map[point(x,y)]==2) que[++tail]=point(x,y);
	}
	while(head<tail){
		head++;
		x=que[head].x;
		y=que[head].y;
		x=find(x),y=find(y);
		V[x].erase(y);
		V[y].erase(x);
		if(V[x].size()>V[y].size()) swap(x,y);
		for(it=V[x].begin();it!=V[x].end();it++){
			int t=*it;
			Map[point(x,t)]=Map[point(t,x)]=0;
			V[y].insert(t);
			V[t].erase(x);
			V[t].insert(y);
			Map[point(y,t)]++;
			Map[point(t,y)]++;
			if(Map[point(y,t)]==2) que[++tail]=point(y,t);
		}
		father[x]=y;
		V[x].clear();
	}
	for(int i=1;i<=n;i++) ans+=(i==father[i]);
	if(ans==1) printf("YES");
	else printf("NO");
	return 0;
}
### Prompt

Please provide a cpp coded solution 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
#pragma GCC optimize(2)
#include <cstdio>
#include <vector>
#include <map>
#include <set>
#define N 1000011
using namespace std;
struct point{
	int x,y;
	point(){}
	point(int xx,int yy){x=xx;y=yy;}
	bool operator<(const point &other)const{
		if(x==other.x) return y<other.y;
		return x<other.x;
	}
}que[N];
set<int> V[N];
set<int>::iterator it;
map<point,int> Map;
int father[N],head,tail,n,ans;
inline int read(){
	int ret=0;char ch=getchar();
	while(ch<'0' || ch>'9') ch=getchar();
	while(ch>='0'&&ch<='9'){ret=ret*10+ch-'0';ch=getchar();}
	return ret;
}
int find(int u){
	if(u==father[u]) return u;
	father[u]=find(father[u]);
	return father[u];
}
int main(){
	int x,y;
	scanf("%d",&n);
	for(int i=1;i<=n;i++) father[i]=i;
	for(int i=1;i<=n*2-2;i++){
		x=read(),y=read();
		Map[point(x,y)]++;
		Map[point(y,x)]++;
		V[x].insert(y);
		V[y].insert(x);
		if(Map[point(x,y)]==2) que[++tail]=point(x,y);
	}
	while(head<tail){
		head++;
		x=que[head].x;
		y=que[head].y;
		x=find(x),y=find(y);
		V[x].erase(y);
		V[y].erase(x);
		if(V[x].size()>V[y].size()) swap(x,y);
		for(it=V[x].begin();it!=V[x].end();it++){
			int t=*it;
			Map[point(x,t)]=Map[point(t,x)]=0;
			V[y].insert(t);
			V[t].erase(x);
			V[t].insert(y);
			Map[point(y,t)]++;
			Map[point(t,y)]++;
			if(Map[point(y,t)]==2) que[++tail]=point(y,t);
		}
		father[x]=y;
		V[x].clear();
	}
	for(int i=1;i<=n;i++) ans+=(i==father[i]);
	if(ans==1) printf("YES");
	else printf("NO");
	return 0;
}
```