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.
# https://codeforces.com/problemset/problem/1067/B
def push(d, u, v):
    if u not in d:
        d[u] = []
    if v not in d:
        d[v] = []
    d[u].append(v)
    d[v].append(u)

def push_v(d, u, val):
    if u not in d:
        d[u] = 0
    d[u] += val
    
n, k = map(int, input().split())
g    = {}

for _ in  range(n-1):
    u, v = map(int, input().split())
    push(g, u, v)
    
deg1 = []
used = [0] * (n+1)

for u in g:
    if len(g[u]) == 1:
        used[u]  = 1
        deg1.append(u)

flg = True        
while k > 0:
    if k >= 1 and len(deg1) < 3:
        flg=False
        break
    
    cnt = {}
    for u in deg1:
        for v in g[u]:
            if used[v] == 0:
                push_v(cnt, v, 1)
    
    for v in deg1:
        used[v] = 1
        
    deg1 = []
    
    for v, val in cnt.items():
        if val < 3:
            flg=False
            break
        deg1.append(v)
    
    if flg==False:
        break
    k-=1
    
if flg==True and len(deg1) > 1:
    flg=False
    
if flg==False:
    print('NO')    
else:
    print('YES')
    