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

def bfs(G, s):
    # la cola comienza con el vertice desde el cual hacemos bfs
    Q = deque() 
    Q.append(s) 
    
    # al inicio todas las distancias comienzan en infinito, por las 
    # restricciones del problema ningun camino va a ser de ese tamanno
    infinite = 10 ** 6
    d = [infinite]*n 

    parent = [-1]*n
    valid = True

    # la distancia del vertice raiz es 0
    d[s] = 0 

    while Q:
        # visitamos u
        u = Q.popleft() 

        not_visited_count = 0

        # visitamos cada adyacente de u
        for v in G[u]: 
            # si no lo hemos visitado, le ponemos distancia y 
            # lo agregamos a la cola para visitar sus adyacentes
            if d[v] == infinite:
                d[v] = d[u] + 1
                parent[v] = u
                Q.append(v)
                not_visited_count += 1
        
        if not_visited_count < 3 and d[u] != k:
            valid = False

    # retornamos el array d, que es el de las distancias del 
    # nodo s al resto de los nodos del grafo            
    return d, parent, valid

leaf = -1
for i,v in enumerate(graph):
    if len(v) == 1:
        leaf = i
        break

d, parent, _ = bfs(graph,leaf)
center = -1
farthest_leaf = -1
path = 2*k

for i,level in enumerate(d):
    if level == path:
        farthest_leaf = i
        break
    
if len(graph[farthest_leaf]) != 1 or farthest_leaf == -1:
    print("NO")
    exit()


for _ in range(k):
    center = parent[farthest_leaf]
    farthest_leaf = center

_, _, valid = bfs(graph,center)

if valid:
    print("YES")
else:
    print("NO")

 




