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, raw_input().strip().split())
children = [[] for i in range(n)]
for i in range(n-1):
    a,b = map(int, raw_input().strip().split())
    children[a-1].append(b-1)
    children[b-1].append(a-1)
leaves = {}
used = [False for i in range(n)]
for i in range(n):
    if len(children[i]) == 1:
        leaves[i] = 1
can = True
for i in range(k):
    newleaves = {}
    for i in leaves.keys():
        count =0
        for j in range(len(children[i])):
            if used[children[i][j]] == False:
                count +=1
                newleaves[children[i][j]] = newleaves.get(children[i][j],0) + 1
        if count > 1 or count == 0:
            can = False
        used[i] = True
    if can == False:
        break
    leaves = newleaves
    for i in leaves.keys():
        if leaves[i] < 3:
            can = False
            break
    if can == False:
        break
if len(leaves.keys()) == 1 and can:
    print 'Yes'
else:
    print 'No'