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.
from collections import defaultdict

n, k = map(int, input().split())

connections = defaultdict(set)

for _ in range(n-1):
	u, v = map(int, input().split())
	connections[u].add(v)
	connections[v].add(u)

leafs = set()
for node in connections:
	if len(connections[node])==1:
		leafs.add(node)

steps = 0
is_correct = True
while is_correct and steps<=k:
	new_leafs = set()
	for x in leafs:
		if len(connections[x])>1:
			is_correct = False
			#print("Len of %d more than one"%x)
			break
		root = list(connections[x])[0]
		if len(connections[root])<4 and len(leafs)!=3:
			is_correct = False
			#print("x: %d Len of root %d less than three"%(x,root))
			#print(connections[root])
			break
	if not is_correct:
		break
	for x in leafs:
		root = list(connections[x])[0]
		new_leafs.add(root)
		connections[root].remove(x)
	leafs = new_leafs
	steps += 1
	if len(leafs)==1 and len(connections[list(leafs)[0]])==0:
		break

#print("steps is %d"%steps)
if is_correct and steps==k:
	print("Yes")
else:
	print('No')
