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 sys import stdin
from collections import *


class graph:
    # initialize graph
    def __init__(self, gdict=None):
        self.gdict, self.l = [[] for _ in range(n + 1)], [0] * (n + 1)

    # add edge
    def addEdge(self, node1, node2, w=None):
        self.gdict[node1].append(node2)
        self.gdict[node2].append(node1)
        self.l[node1] += 1
        self.l[node2] += 1

    def bfs_util(self, i):
        self.visit, self.diam = [0] * (n + 1), [0] * (n + 1)
        queue, self.visit[i], self.gdict2 = deque([i]), 1, [0 for _ in range(n + 1)]

        while queue:
            # dequeue parent vertix
            s = queue.popleft()

            # enqueue child vertices
            for i in self.gdict[s]:
                if self.visit[i] == 0:
                    queue.append(i)
                    self.visit[i] = 1
                    self.diam[i] += self.diam[s] + 1
                    self.gdict2[i] = s

        return self.diam.index(max(self.diam))

    def diameter(self, v):
        first = self.bfs_util(v)
        path = [first]
        while first != v:
            first = self.gdict2[first]
            path.append(first)

        if k >= len(path) or g.l[path[k]] < 3:
            print('No')
            exit()
        g.bfs_util(path[k])

        for i in range(1, n + 1):
            if (2 <= g.l[i] <= 3 and i != path[k]) or (g.l[i] == 1 and self.diam[i] != k):
                print('No')
                exit()


rints = lambda: [int(x) for x in stdin.readline().split()]
n, k = rints()
g = graph()

for _ in range(n - 1):
    u, v = rints()
    g.addEdge(u, v)

if n == 1:
    print('NO')
    exit()

g.diameter(g.l.index(1))
print('Yes')
