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.
import sys
range = xrange
input = sys.stdin.readline

n,k = [int(x) for x in input().split()]

coupl = [[] for _ in range(n)]

inp = [int(x)-1 for line in sys.stdin for x in line.split()]
ii = 0

for _ in range(n-1):
    a,b = inp[ii],inp[ii+1]
    ii+=2
    coupl[a].append(b)
    coupl[b].append(a)

if not any(len(coupl[i])>=3 for i in range(n)):
    print 'No'
    sys.exit()

Q = []
found = [False]*n

for i in range(n):
    if len(coupl[i])==2:
        print 'No'
        sys.exit()
    
    if len(coupl[i])==1:
        Q.append(i)
        found[i] = True

count = len(Q)
counter = [0]*n

big_count = 0
times = 0
while count<n:
    times += 1
    big_count = count
    new_Q = []
    for node in Q:
        for nei in coupl[node]:
            if not found[nei]:
                
                if counter[nei]==0:
                    new_Q.append(nei)
                
                counter[nei] += 1
        
    for node in new_Q:
        if counter[node]<3 or (counter[node]!=len(coupl[node])-1 and big_count!=n-1) or (counter[node]!=len(coupl[node]) and big_count==n-1):
            print 'No'
            sys.exit()
        found[node] = True
    
    Q = new_Q
    if len(Q)==0:
        print 'No'
        sys.exit()
    count += len(Q)

if big_count==n-1 and times==k:
    print 'Yes'
else:
    print 'No'
