#include<bits/stdc++.h>

using namespace std;

typedef pair<int,int> pii;
#define N 300000
#define P(x,y) make_pair(x,y)

int n;
multiset<int> s[N];
map<int,int> mp[N];
queue<pii> q;

void ins(int x,int y){
	if (x==y) return;
	++mp[x][y]; ++mp[y][x];
	if (mp[x][y]==2) q.push(P(x,y));
	s[x].insert(y); s[y].insert(x);
}

int main(){
	scanf("%d",&n);
	for (int i=1;i<n*2-1;++i){
		int x,y; scanf("%d%d",&x,&y);
		ins(x,y);
	}
	for (int i=1;i<n;++i){
		while (1){
			if (q.empty()){puts("NO"); return 0;}
			int x=q.front().first,y=q.front().second; q.pop();
			if (!mp[x][y]) continue;
			if (s[x].size()<s[y].size()) swap(x,y);
			for (int i:s[y]){
				mp[y][i]=mp[i][y]=0;
				s[i].erase(s[i].find(y));
				ins(i,x);
			}
			break;
		}
	}
	puts("YES");
	
	return 0;
}
			
### 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>

using namespace std;

typedef pair<int,int> pii;
#define N 300000
#define P(x,y) make_pair(x,y)

int n;
multiset<int> s[N];
map<int,int> mp[N];
queue<pii> q;

void ins(int x,int y){
	if (x==y) return;
	++mp[x][y]; ++mp[y][x];
	if (mp[x][y]==2) q.push(P(x,y));
	s[x].insert(y); s[y].insert(x);
}

int main(){
	scanf("%d",&n);
	for (int i=1;i<n*2-1;++i){
		int x,y; scanf("%d%d",&x,&y);
		ins(x,y);
	}
	for (int i=1;i<n;++i){
		while (1){
			if (q.empty()){puts("NO"); return 0;}
			int x=q.front().first,y=q.front().second; q.pop();
			if (!mp[x][y]) continue;
			if (s[x].size()<s[y].size()) swap(x,y);
			for (int i:s[y]){
				mp[y][i]=mp[i][y]=0;
				s[i].erase(s[i].find(y));
				ins(i,x);
			}
			break;
		}
	}
	puts("YES");
	
	return 0;
}
			
```