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 p[100000];
int find(int x) { return x == p[x] ? x : (p[x] = find(p[x])); }
void unite(int a, int b) {
  if (a % 2 == 0) swap(a, b);
  p[find(a)] = find(b);
}
int main(int argc, char *argv[]) {
  ios::sync_with_stdio(false);
  int N, K;
  cin >> N >> K;
  vector<set<int>> g(N);
  for (int i = 1; i < N; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].insert(b);
    g[b].insert(a);
  }
  int ans = 0, pans = 0;
  while (K--) {
    iota(begin(p), begin(p) + N, 0);
    for (int i = 0; i < g.size(); i++)
      for (int j : g[i]) unite(i, j);
    int t = -1;
    for (int i = 0; i < N; i++)
      if (find(i) != i) t = find(i);
    if (t < 0) {
      cout << "No" << endl;
      return 0;
    }
    for (int i = 0; i < N; i++)
      if (find(i) != t and find(i) != i) {
        cout << "No" << endl;
        return 0;
      }
    vector<size_t> sz;
    for (int i = 0; i < N; i++) sz.push_back(g[i].size());
    vector<int> cnt(N, 0);
    for (int i = 0; i < N; i++) {
      if (sz[i] == 2) {
        cout << "No" << endl;
        return 0;
      } else if (sz[i] == 1) {
        cnt[*g[i].begin()]++;
        g[*g[i].begin()].erase(i);
        g[i].clear();
        ans++;
      }
    }
    for (int i : cnt)
      if (i < 3 and i > 0) {
        cout << "No" << endl;
        return 0;
      }
    if (ans - pans < 3) {
      cout << "No" << endl;
      return 0;
    }
    pans = ans;
  }
  cout << (ans == N - 1 ? "Yes" : "No") << endl;
  return 0;
}
