Each New Year Timofey and his friends cut down a tree of *n* vertices and bring it home. After that they paint all the *n* its vertices, so that the *i*-th vertex gets color *c*<sub class="lower-index">*i*</sub>.

Now it's time for Timofey birthday, and his mother asked him to remove the tree. Timofey removes the tree in the following way: he takes some vertex in hands, while all the other vertices move down so that the tree becomes rooted at the chosen vertex. After that Timofey brings the tree to a trash can.

Timofey doesn't like it when many colors are mixing together. A subtree annoys him if there are vertices of different color in it. Timofey wants to find a vertex which he should take in hands so that there are no subtrees that annoy him. He doesn't consider the whole tree as a subtree since he can't see the color of the root vertex.

A subtree of some vertex is a subgraph containing that vertex and all its descendants.

Your task is to determine if there is a vertex, taking which in hands Timofey wouldn't be annoyed.
The first line contains single integer *n* (2 ≤ *n* ≤ 10<sup class="upper-index">5</sup>) — the number of vertices in the tree.

Each of the next *n* - 1 lines contains two integers *u* and *v* (1 ≤ *u*, *v* ≤ *n*, *u* ≠ *v*), denoting there is an edge between vertices *u* and *v*. It is guaranteed that the given graph is a tree.

The next line contains *n* integers *c*<sub class="lower-index">1</sub>, *c*<sub class="lower-index">2</sub>, ..., *c*<sub class="lower-index">*n*</sub> (1 ≤ *c*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">5</sup>), denoting the colors of the vertices.
none
4
1 2
2 3
3 4
1 2 1 1
YES
2
3
1 2
2 3
1 2 3
YES
2
4
1 2
2 3
3 4
1 2 1 2
NO
3
2 1
2 3
1 2 3
YES
2
4
1 2
2 4
4 3
1 1 3 2
YES
4
2
1 2
1 1
YES
1
10
5 7
4 5
10 2
3 6
1 2
3 4
8 5
4 9
2 3
15 15 15 15 5 15 26 18 15 15
YES
5
8
1 2
1 3
3 5
3 6
1 4
4 7
4 8
1 3 1 1 1 1 1 2
NO
3
2 1
2 3
4 4 4
YES
1
3
1 2
1 3
1 2 2
YES
1
4
1 4
2 4
3 4
1 2 3 1
YES
4
4
1 2
1 3
1 4
1 2 3 4
YES
1
9
1 2
2 3
3 4
4 5
2 7
7 6
2 8
8 9
1 1 2 2 2 3 3 4 4
YES
2
3
2 1
2 3
4 4 5
YES
2
4
1 2
2 3
3 4
1 2 2 1
NO
class Graph:
	def __init__(self,v):
		self.v = v
		self.array = [ [] for _ in range(v)]
		self.visited = [False for _ in range(v)]
		self.color = [0 for _ in range(v)]
	
	def addEdge(self,u,v):
		self.array[u].append(v)
		self.array[v].append(u)

def isSameUtil(g,src,color):
	g.visited[src] = True
	if not g.color[src] == color:
		return False
	
	ret = True
	for i in range(len(g.array[src])):
		child = g.array[src][i]
		if g.visited[child]:
			continue
		
		ret = ret and isSameUtil(g,g.array[src][i],color)
	return ret

def isSame(g,src):
	return isSameUtil(g,src,g.color[src])
	
def isFine(g,src):	
	g.visited = [False for _ in range(g.v)]
	ret = True
	g.visited[src] = True
	for i in range(len(g.array[src])):
		ret = ret and isSame(g,g.array[src][i])
		
	return ret

def solve():
	n = int(input())
	g = Graph(n)
	for i in range(n-1):
		u,v = list(map(int,input().split()))
		g.addEdge(u-1,v-1)
	g.color = list(map(int,input().split()))

	done = False
	for u in range(n):
		for j in range(len(g.array[i])):
			v = g.array[u][j]
			if not g.color[u] == g.color[v]:	
				if isFine(g,u):
					print("YES")
					print(str(u+1))
					done = True
					break
				elif isFine(g,v):
					print("YES")
					print(str(v+1))
					done = True
					break
				else:
					print("NO")
					done = True
					break
		if done:
			break

if __name__ == "__main__":
	solve()