#include<bits/stdc++.h>
using namespace std;

const int _ = 2e5 + 7;
int fa[_] , N , S[_] , T[_];
set < int > node[_]; map < pair < int , int > , int > ext;
queue < pair < int , int > > q;
void del(int x){node[S[x]].erase(x); node[T[x]].erase(x);}
void add(int x){
	if(S[x] > T[x]) swap(S[x] , T[x]);
	pair < int , int > t(S[x] , T[x]);
	if(ext.find(t) == ext.end()) ext[t] = x;
	else{del(x); auto p = ext.find(t); del(p->second); q.push(t); ext.erase(p);}
}

int find(int x){return fa[x] == x ? x : (fa[x] = find(fa[x]));}
void merge(int p , int q){
	p = find(p); q = find(q); if(node[p].size() < node[q].size()) swap(p , q);
	fa[q] = p;
	for(auto t : node[q]){
		ext.erase(make_pair(S[t] , T[t]));
		node[p].insert(t); (S[t] == q ? S[t] : T[t]) = p; add(t);
	}
}

int main(){
	cin >> N; for(int i = 1 ; i <= N ; ++i) fa[i] = i;
	for(int i = 1 ; i <= 2 * N - 2 ; ++i){int A , B; cin >> A >> B; node[A].insert(i); node[B].insert(i); S[i] = A; T[i] = B;}
	for(int i = 1 ; i <= 2 * N - 2 ; ++i) add(i);
	
	while(!q.empty()){int x = q.front().first , y = q.front().second; q.pop(); merge(x , y);}
	cout << (ext.empty() ? "YES" : "NO");
	return 0;
}

### Prompt

Your task is to create a CPP solution 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 _ = 2e5 + 7;
int fa[_] , N , S[_] , T[_];
set < int > node[_]; map < pair < int , int > , int > ext;
queue < pair < int , int > > q;
void del(int x){node[S[x]].erase(x); node[T[x]].erase(x);}
void add(int x){
	if(S[x] > T[x]) swap(S[x] , T[x]);
	pair < int , int > t(S[x] , T[x]);
	if(ext.find(t) == ext.end()) ext[t] = x;
	else{del(x); auto p = ext.find(t); del(p->second); q.push(t); ext.erase(p);}
}

int find(int x){return fa[x] == x ? x : (fa[x] = find(fa[x]));}
void merge(int p , int q){
	p = find(p); q = find(q); if(node[p].size() < node[q].size()) swap(p , q);
	fa[q] = p;
	for(auto t : node[q]){
		ext.erase(make_pair(S[t] , T[t]));
		node[p].insert(t); (S[t] == q ? S[t] : T[t]) = p; add(t);
	}
}

int main(){
	cin >> N; for(int i = 1 ; i <= N ; ++i) fa[i] = i;
	for(int i = 1 ; i <= 2 * N - 2 ; ++i){int A , B; cin >> A >> B; node[A].insert(i); node[B].insert(i); S[i] = A; T[i] = B;}
	for(int i = 1 ; i <= 2 * N - 2 ; ++i) add(i);
	
	while(!q.empty()){int x = q.front().first , y = q.front().second; q.pop(); merge(x , y);}
	cout << (ext.empty() ? "YES" : "NO");
	return 0;
}

```