#include "bits/stdc++.h"
using namespace std;
const int N = 1e5 + 5;
int n;
int a[N] , b[N];
int c[N] , d[N];
map < int , int > edges[N];
queue < pair < int , int > > q;
int comp[N];
int find(int node){
	if(comp[node] == node){
		return node;
	}
	return comp[node] = find(comp[node]);
}
void insert(int a , int b){
	a = find(a);
	b = find(b);
	if(a != b){
		++edges[a][b];
		++edges[b][a];
		if(edges[a][b] >= 2){
			q.push({a , b});
		}
	}
}
void merge(){
	int a = - 1 , b = -1;
	while(!q.empty()){
		a = find(q.front().first);
		b = find(q.front().second);
		q.pop();
		if(a != b){
			break;
		}
	}
	if(a == b){
		printf("NO\n");
		exit(0);
	}
	if(edges[a].size() > edges[b].size()){
		swap(a , b);
	}
	for(auto it : edges[a]){
		edges[it.first].erase(edges[it.first].find(a));
	}
	comp[a] = b;
	for(auto it : edges[a]){
		if(it.first == b){
			continue;
		}
		if(it.first == a){
			continue;
		}
		while(it.second--){
			insert(b , find(it.first));
		}
	}
	edges[a].clear();
}
int main(){
	scanf("%d" , &n);
	for(int i = 1 ; i < n ; ++i){
		scanf("%d %d" , a + i , b + i);
	}
	for(int i = 1 ; i < n ; ++i){
		scanf("%d %d" , c + i , d + i);
	}
	for(int i = 1 ; i <= n ; ++i){
		comp[i] = i;
	}
	for(int i = 1 ; i < n ; ++i){
		insert(a[i] , b[i]);
		insert(c[i] , d[i]);
	}
	for(int i = 1 ; i < n ; ++i){
		merge();
	}
	printf("YES\n");
}
### 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 N = 1e5 + 5;
int n;
int a[N] , b[N];
int c[N] , d[N];
map < int , int > edges[N];
queue < pair < int , int > > q;
int comp[N];
int find(int node){
	if(comp[node] == node){
		return node;
	}
	return comp[node] = find(comp[node]);
}
void insert(int a , int b){
	a = find(a);
	b = find(b);
	if(a != b){
		++edges[a][b];
		++edges[b][a];
		if(edges[a][b] >= 2){
			q.push({a , b});
		}
	}
}
void merge(){
	int a = - 1 , b = -1;
	while(!q.empty()){
		a = find(q.front().first);
		b = find(q.front().second);
		q.pop();
		if(a != b){
			break;
		}
	}
	if(a == b){
		printf("NO\n");
		exit(0);
	}
	if(edges[a].size() > edges[b].size()){
		swap(a , b);
	}
	for(auto it : edges[a]){
		edges[it.first].erase(edges[it.first].find(a));
	}
	comp[a] = b;
	for(auto it : edges[a]){
		if(it.first == b){
			continue;
		}
		if(it.first == a){
			continue;
		}
		while(it.second--){
			insert(b , find(it.first));
		}
	}
	edges[a].clear();
}
int main(){
	scanf("%d" , &n);
	for(int i = 1 ; i < n ; ++i){
		scanf("%d %d" , a + i , b + i);
	}
	for(int i = 1 ; i < n ; ++i){
		scanf("%d %d" , c + i , d + i);
	}
	for(int i = 1 ; i <= n ; ++i){
		comp[i] = i;
	}
	for(int i = 1 ; i < n ; ++i){
		insert(a[i] , b[i]);
		insert(c[i] , d[i]);
	}
	for(int i = 1 ; i < n ; ++i){
		merge();
	}
	printf("YES\n");
}
```