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 deque
n,k = map(int,input().split())
d = {}
for i in range(n):
    d[i+1] = set()
for i in range(n-1):
    u,v = map(int,input().split())
    d[u].add(v)
    d[v].add(u)
dist = {}
prev = {}
dist[1] = 0
q = deque()
q.append(1)
while len(q) >0:
    cur =  q.popleft()
    for x in d[cur]:
        if x not in dist:
            prev[x] = cur
            dist[x] = dist[cur]+1
            q.append(x)
answer = True
if k > dist[cur]:
    answer = False
else:
    for i in range(k):
        cur = prev[cur]
    dist2 = {}
    dist2[cur] = 0
    q = deque()
    q.append(cur)
    while len(q) >0:
        cur2 =  q.popleft()
        if cur2 == cur and len(d[cur]) < 3:
            answer = False
            break
        if len(d[cur2]) == 1:
            if dist2[cur2] != k:
                answer = False
                break
        elif len(d[cur2]) < 4 and cur2 != cur:
            answer = False
            break
        for x in d[cur2]:
            if x not in dist2:
                dist2[x] = dist2[cur2]+1
                q.append(x)
if answer:
    print("Yes")
else:
    print("No")
