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 sys import stdin
from collections import deque

n, k = map(int, stdin.readline().split())

graph = [set() for _ in range(n)]

for _ in range(n-1):
    a,b = map(int,stdin.readline().split())

    graph[a - 1].add(b - 1)
    graph[b - 1].add(a - 1)

leafs = [i for i,v in enumerate(graph) if len(v) == 1]
new_leafs = []
valid = True
centers = dict()
count = 0

while len(leafs) > 1 and valid:
    for leaf in leafs:
        center = graph[leaf].pop()

        try:
            centers[center] += 1
        except KeyError:
            centers[center] = 1

        graph[center].remove(leaf)

        if len(graph[center]) == 0:
            break

        elif len(graph[center]) == 1:
            new_leafs.append(center)

    if any(mult < 3 for mult in centers.values()):
        valid = False
        break
    
    count = count + 1
    leafs = new_leafs
    new_leafs = []
    centers = {}

if valid and count == k:
    print("YES")
else:
    print("NO")
