#include<bits/stdc++.h>
#define fi first
#define sc second
const int N=100010;
using namespace std;
int n,fa[N];
multiset<int> s[N];
map<pair<int,int> ,int> mps1,mps2;
queue<pair<int,int> >q;
inline pair<int,int> makepair(int x,int y){
	if(x>y)return pair<int,int>(y,x);
	return pair<int,int>(x,y);
}
inline int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}
inline int read(){
	int f=1,x=0;char ch;
	do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
	do{x=x*10+ch-'0';ch=getchar();}while(ch>='0'&&ch<='9');
	return f*x;
}
int main(){
	n=read();
	for(int i=1;i<=n;i++)fa[i]=i;
	for(int i=1;i<2*n-1;i++){
		int u=read(),v=read();
		s[u].insert(v);s[v].insert(u);
		pair<int,int> p=makepair(u,v);
		if((++mps1[p]>1)&&!mps2.count(p))mps2[p]=1,q.push(p);
	}
	for(int i=1;i<n;i++){
		int x,y;
		while(1){
			if(q.empty())return puts("NO"),0;
			x=find(q.front().fi);y=find(q.front().sc);q.pop();
			if(x!=y)break;
		}
		if(s[x].size()>s[y].size()) swap(x,y); fa[x]=y;
        for(set<int>::iterator it=s[x].begin();it!=s[x].end();it++){
            int v=find(*it);
            if(v==y)continue;
            s[y].insert(v); s[v].insert(y);
            if((++mps1[makepair(y,v)])>1&&!mps2.count(makepair(y,v)))
                q.push(makepair(y,v)),mps2[makepair(y,v)]=1;
            s[v].erase(s[v].find(x));
        }
        s[y].erase(x);s[x].clear();
	}
	puts("YES");
}
### 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<bits/stdc++.h>
#define fi first
#define sc second
const int N=100010;
using namespace std;
int n,fa[N];
multiset<int> s[N];
map<pair<int,int> ,int> mps1,mps2;
queue<pair<int,int> >q;
inline pair<int,int> makepair(int x,int y){
	if(x>y)return pair<int,int>(y,x);
	return pair<int,int>(x,y);
}
inline int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}
inline int read(){
	int f=1,x=0;char ch;
	do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
	do{x=x*10+ch-'0';ch=getchar();}while(ch>='0'&&ch<='9');
	return f*x;
}
int main(){
	n=read();
	for(int i=1;i<=n;i++)fa[i]=i;
	for(int i=1;i<2*n-1;i++){
		int u=read(),v=read();
		s[u].insert(v);s[v].insert(u);
		pair<int,int> p=makepair(u,v);
		if((++mps1[p]>1)&&!mps2.count(p))mps2[p]=1,q.push(p);
	}
	for(int i=1;i<n;i++){
		int x,y;
		while(1){
			if(q.empty())return puts("NO"),0;
			x=find(q.front().fi);y=find(q.front().sc);q.pop();
			if(x!=y)break;
		}
		if(s[x].size()>s[y].size()) swap(x,y); fa[x]=y;
        for(set<int>::iterator it=s[x].begin();it!=s[x].end();it++){
            int v=find(*it);
            if(v==y)continue;
            s[y].insert(v); s[v].insert(y);
            if((++mps1[makepair(y,v)])>1&&!mps2.count(makepair(y,v)))
                q.push(makepair(y,v)),mps2[makepair(y,v)]=1;
            s[v].erase(s[v].find(x));
        }
        s[y].erase(x);s[x].clear();
	}
	puts("YES");
}
```