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.
#include <bits/stdc++.h>
using namespace std;
int const N = 100000;
vector<int> tr[N + 1];
int n, k;
void no() {
  printf("No\n");
  exit(0);
}
pair<int, int> getf(int v, int p = 0, int d = 0) {
  pair<int, int> an(d, v);
  for (int i = 0; i < (int)(tr[v].size()); ++i) {
    int u = tr[v][i];
    if (u == p) continue;
    an = max(an, getf(u, v, d + 1));
  }
  return an;
}
int getm(int v, int ds, int mxd, int p = 0, int d = 0) {
  if (v == ds) return v;
  for (int i = 0; i < (int)(tr[v].size()); ++i) {
    int u = tr[v][i];
    if (u == p) continue;
    int t = getm(u, ds, mxd, v, d + 1);
    if (t) {
      if (d >= mxd >> 1) return v;
      return t;
    }
  }
  return 0;
}
void go(int v, int p = 0, int d = 0) {
  if (d == k) {
    if ((int)tr[v].size() != 1) no();
    return;
  }
  if ((int)tr[v].size() < 4 - (p == 0)) no();
  for (int i = 0; i < (int)(tr[v].size()); ++i) {
    int u = tr[v][i];
    if (u == p) continue;
    go(u, v, d + 1);
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < (int)(n); ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    tr[a].push_back(b);
    tr[b].push_back(a);
  }
  int a = getf(1).second;
  pair<int, int> b = getf(a);
  if (b.first != k << 1) no();
  a = getm(a, b.second, b.first);
  go(a);
  printf("Yes\n");
}
