#include <bits/stdc++.h>
using namespace std;
const int N=1000005;
typedef set<int>::iterator It;
set<int> S[N];
queue<pair<int,int> > q;
map<pair<int,int>,int> Map;
int FA[N],n;
int find(int u) { if (u!=FA[u]) FA[u]=find(FA[u]); return FA[u]; }
pair<int,int> work(int u,int v) { if (u>v) swap(u,v); return make_pair(u,v); }
void Add(int u,int v) {
	if (u>v) swap(u,v);
	S[u].insert(v),S[v].insert(u);
	Map[work(u,v)]++; int num=Map[work(u,v)];
	if (num==2) q.push(work(u,v));
}
void Del(int u,int v) {
	S[u].erase(v); S[v].erase(u);
	Map.erase(work(u,v));
}
bool Solve() {
	for (int i=1; i<=n; i++) FA[i]=i;
	for (int i=1; i<n; i++) {
		if (q.empty()) return false;
		int u=q.front().first,v=q.front().second;
		q.pop(); u=find(u),v=find(v);
		if (S[u].size()>S[v].size()) swap(u,v);
		FA[u]=v; Del(u,v);
		for (It i=S[u].begin(),j; i!=S[u].end(); i=j) {
			(j=i)++; int x=*i; x=find(x);
			Del(u,x); Add(v,x);
		}
	}
	return true;
}
int main() {

	scanf("%d",&n);
	for (int i=1; i<=(n-1)<<1; i++) {
		int u,v; scanf("%d%d",&u,&v);
		Add(u,v);
	}
	if (Solve()) printf("YES\n"); else printf("NO\n");
	return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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 N=1000005;
typedef set<int>::iterator It;
set<int> S[N];
queue<pair<int,int> > q;
map<pair<int,int>,int> Map;
int FA[N],n;
int find(int u) { if (u!=FA[u]) FA[u]=find(FA[u]); return FA[u]; }
pair<int,int> work(int u,int v) { if (u>v) swap(u,v); return make_pair(u,v); }
void Add(int u,int v) {
	if (u>v) swap(u,v);
	S[u].insert(v),S[v].insert(u);
	Map[work(u,v)]++; int num=Map[work(u,v)];
	if (num==2) q.push(work(u,v));
}
void Del(int u,int v) {
	S[u].erase(v); S[v].erase(u);
	Map.erase(work(u,v));
}
bool Solve() {
	for (int i=1; i<=n; i++) FA[i]=i;
	for (int i=1; i<n; i++) {
		if (q.empty()) return false;
		int u=q.front().first,v=q.front().second;
		q.pop(); u=find(u),v=find(v);
		if (S[u].size()>S[v].size()) swap(u,v);
		FA[u]=v; Del(u,v);
		for (It i=S[u].begin(),j; i!=S[u].end(); i=j) {
			(j=i)++; int x=*i; x=find(x);
			Del(u,x); Add(v,x);
		}
	}
	return true;
}
int main() {

	scanf("%d",&n);
	for (int i=1; i<=(n-1)<<1; i++) {
		int u,v; scanf("%d%d",&u,&v);
		Add(u,v);
	}
	if (Solve()) printf("YES\n"); else printf("NO\n");
	return 0;
}
```