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 __future__ import division
from fileinput import input
from collections import defaultdict, deque

inp = input()
n, k = map(int, inp.readline().split())

graph = defaultdict(list)
for _ in range(n - 1):
    u, v = map(int, inp.readline().split())
    graph[u].append(v)
    graph[v].append(u)


def bfs(a):
    todo = deque()
    todo.append(a)
    vis = set()
    vis.add(a)
    fro = {}

    while len(todo) > 0:
        cur = todo.popleft()
        last = cur
        for ne in graph[cur]:
            if ne not in vis:
                vis.add(ne)
                todo.append(ne)
                fro[ne] = cur
    return last, fro


v2, _ = bfs(1)
v3, fro = bfs(v2)

path = [v3]
cur = v3
while cur in fro:
    path.append(fro[cur])
    cur = fro[cur]

if len(path) != k*2 + 1:
    print "No"
    exit()
mid = path[len(path) // 2]


def hh(mid, k, p):
    if k == 1:
        count = 0
        for ne in graph[mid]:
            if ne != p:
                count += 1
                if len(graph[ne]) != 1:
                    return False
        return count >= 3
    count = 0
    for ne in graph[mid]:
        if ne != p:
            count += 1
            if not hh(ne, k - 1, mid):
                return False
    return count >= 3


if hh(mid, k, 1000000000000):
    print "Yes"
else:
    print "No"
