p03727 AtCoder Grand Contest 014 - Blue and Red Tree
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
```python
import queue
n,q=int(input()),queue.Queue()
s=[set() for i in range(n+1)]
for i in range(2*n-2):
	u,v=map(int,input().split())
	if v in s[u]:
		q.put((u,v))
	else:
		s[u].add(v)
		s[v].add(u)
f=[i for i in range(n+1)]
def find(x):
	if f[x]==x:
		return x
	else:
		f[x]=find(f[x])
		return f[x]
while not q.empty():
	u,v=map(find,q.get())
	if u==v:
		continue
	if len(s[u])<len(s[v]):
		u,v=v,u
	s[u].remove(v)
	s[v].remove(u)
	for x in s[v]:
		s[x].remove(v)
		if u in s[x]:
			q.put((u,x))
		else:
			s[u].add(x)
			s[x].add(u)
	s[v],f[v]=[],u
root=find(1)
for i in range(2,n+1):
	if find(i)!=root:
		print("NO")
		break
else:
	print("YES")
```