Someone give a strange birthday present to Ivan. It is hedgehog — connected undirected graph in which one vertex has degree at least 3 (we will call it center) and all other vertices has degree 1. Ivan thought that hedgehog is too boring and decided to make himself k-multihedgehog.

Let us define k-multihedgehog as follows:

  * 1-multihedgehog is hedgehog: it has one vertex of degree at least 3 and some vertices of degree 1.
  * For all k ≥ 2, k-multihedgehog is (k-1)-multihedgehog in which the following changes has been made for each vertex v with degree 1: let u be its only neighbor; remove vertex v, create a new hedgehog with center at vertex w and connect vertices u and w with an edge. New hedgehogs can differ from each other and the initial gift. 



Thereby k-multihedgehog is a tree. Ivan made k-multihedgehog but he is not sure that he did not make any mistakes. That is why he asked you to check if his tree is indeed k-multihedgehog.

Input

First line of input contains 2 integers n, k (1 ≤ n ≤ 10^{5}, 1 ≤ k ≤ 10^{9}) — number of vertices and hedgehog parameter.

Next n-1 lines contains two integers u v (1 ≤ u,    v ≤ n;    u ≠ v) — indices of vertices connected by edge.

It is guaranteed that given graph is a tree.

Output

Print "Yes" (without quotes), if given graph is k-multihedgehog, and "No" (without quotes) otherwise.

Examples

Input

14 2
1 4
2 4
3 4
4 13
10 5
11 5
12 5
14 5
5 13
6 7
8 6
13 6
9 6


Output

Yes


Input

3 1
1 3
2 3


Output

No

Note

2-multihedgehog from the first example looks like this:

<image>

Its center is vertex 13. Hedgehogs created on last step are: [4 (center), 1, 2, 3], [6 (center), 7, 8, 9], [5 (center), 10, 11, 12, 13].

Tree from second example is not a hedgehog because degree of center should be at least 3.
n,k = map(int,input().split(" "))
degrees = [0] * n
neighbors = [list() for x in range(n)]
for i in range(n-1):
	first,second = map(int,input().split(" "))
	degrees[first-1] += 1
	degrees[second-1] += 1
	neighbors[first-1] += [second]
	neighbors[second-1] += [first]

# start at a leaf
curr = 0
for i in range(n):
	if degrees[i] == 1:
		curr = i+1
		break
if curr == 0 or len(neighbors[curr-1]) == 0:
	print("No")
	exit()
curr = neighbors[curr-1][0]


def check(prev,parent,curr,level,degrees,neighbors,k):
	#print("curr: ",curr)
	#print("level: ",level)
	if level == 0:
		return len(parent) == 1 and degrees[curr-1] == 1,[]
	checked = []
	for neighbor in neighbors[curr-1]:
		#print("neighbor: ",neighbor)
		#print("checked: ",checked)
		#print("parent: ",parent)
		if len(prev) != 0 and prev[0] == neighbor:
			checked += [neighbor]
			continue
		if len(parent) != 0 and parent[0] == neighbor:
			continue
		result,garbage = check([],[curr],neighbor,level-1,degrees,neighbors,k)
		if result:
			checked += [neighbor]
		else:
			#print("adding the parent")
			if len(parent) == 0:
				parent += [neighbor]
			else:
				return False,[]
	if len(checked) > 2 and len(parent) == 0 and level == k:
		#print("first check")
		return True,[]
	elif len(checked) > 2 and len(parent) == 1 and level != k:
		#print("second check")
		return True,parent
	else:
		#print("len(checked): ",len(checked))
		#print("len(parent): ",len(parent))
		#print("level: ",level)
		#print("the end fail statement")
		return False,[]

prev = []
parent = []
counter = 1
while(counter <= k):
	result,parent = check(prev,[],curr,counter,degrees,neighbors,k)
	if not(result):
		print("No")
		exit()
	if counter == k:
		print("Yes")
		exit()
	prev = [curr]
	curr = parent[0]
	counter += 1

		


