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

Source: atcoder | Difficulty: UNKNOWN_DIFFICULTY


Solutions:
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')