#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+10;
map<int,map<int,int> >RB;
set<int>s[maxn];
set<int>::iterator it;
int qa[maxn],qb[maxn],fa[maxn],siz[maxn],h=1,e;
int get(int x) {
	return x==fa[x]?x:fa[x]=get(fa[x]);
}
int main() {
	int n,x,y;
	scanf("%d",&n);
	for(int i=1;i<2*n-1;i++) {
		scanf("%d%d",&x,&y);
		s[x].insert(y);
		s[y].insert(x);
		if(x>y)swap(x,y);
		RB[x][y]++;
		if(RB[x][y]==2)qa[++e]=x,qb[e]=y;
	}
	for(int i=1;i<=n;i++)fa[i]=i;
	for(int T=1;T<n;T++) {
		if(h>e) {
			puts("NO");
			return 0;
		}
		while(h<=e&&(x=get(qa[h]))&&(y=get(qb[h++]))&&x==y);
		if(x==y) {
			puts("NO");
			return 0;
		}
		if(s[x].size()>s[y].size())swap(x,y);
		fa[x]=y;
		for(it=s[x].begin();it!=s[x].end();it++) {
			int v=*s[x].find(*it),u=y;
			s[v].erase(s[v].find(x));
			if(v==u)continue;
			s[v].insert(u);
			s[u].insert(v);
			if(v>u)swap(v,u);
			RB[v][u]++;
			if(RB[v][u]==2)qa[++e]=v,qb[e]=u;
		}
	}
	puts("YES");
	return 0;
}
### 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>
using namespace std;
const int maxn=1e5+10;
map<int,map<int,int> >RB;
set<int>s[maxn];
set<int>::iterator it;
int qa[maxn],qb[maxn],fa[maxn],siz[maxn],h=1,e;
int get(int x) {
	return x==fa[x]?x:fa[x]=get(fa[x]);
}
int main() {
	int n,x,y;
	scanf("%d",&n);
	for(int i=1;i<2*n-1;i++) {
		scanf("%d%d",&x,&y);
		s[x].insert(y);
		s[y].insert(x);
		if(x>y)swap(x,y);
		RB[x][y]++;
		if(RB[x][y]==2)qa[++e]=x,qb[e]=y;
	}
	for(int i=1;i<=n;i++)fa[i]=i;
	for(int T=1;T<n;T++) {
		if(h>e) {
			puts("NO");
			return 0;
		}
		while(h<=e&&(x=get(qa[h]))&&(y=get(qb[h++]))&&x==y);
		if(x==y) {
			puts("NO");
			return 0;
		}
		if(s[x].size()>s[y].size())swap(x,y);
		fa[x]=y;
		for(it=s[x].begin();it!=s[x].end();it++) {
			int v=*s[x].find(*it),u=y;
			s[v].erase(s[v].find(x));
			if(v==u)continue;
			s[v].insert(u);
			s[u].insert(v);
			if(v>u)swap(v,u);
			RB[v][u]++;
			if(RB[v][u]==2)qa[++e]=v,qb[e]=u;
		}
	}
	puts("YES");
	return 0;
}
```