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;
signed main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  map<long long, vector<long long> > graph;
  long long deg[(long long)1e5 + 1] = {0}, d3 = 0;
  cin >> n >> k;
  long long p = -1;
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    graph[u].push_back(v);
    graph[v].push_back(u);
    deg[v]++;
    deg[u]++;
  }
  bool visited[(long long)1e5 + 1] = {false};
  unordered_set<long long> deg1;
  for (long long i = 1; i <= n; ++i) {
    if (deg[i] == 1) {
      deg1.insert(i);
      visited[i] = true;
    }
    if (deg[i] == 2) {
      cout << "No";
      return 0;
    }
    if (deg[i] == 3) d3++;
  }
  if (d3 > 1) {
    cout << "No";
    return 0;
  }
  for (long long first = 0; first < k; ++first) {
    if (deg1.size() == 1) {
      cout << "No" << '\n';
      return 0;
    }
    auto deg2 = deg1;
    for (auto i : deg1) {
      for (long long j : graph[i]) {
        if (visited[j]) continue;
        deg[j]--;
        if (deg[j] == 1) {
          deg2.insert(j);
          visited[j] = true;
          if (graph[j].size() < 4 &&
              !(graph[j].size() == 3 && first == k - 1)) {
            cout << "No";
            return 0;
          }
          if (graph[j].size() == 3 && first == k - 1) d3 = -1;
        }
        break;
      }
      deg2.erase(i);
    }
    deg1 = deg2;
  }
  if (deg1.size() == 1 &&
      !((n == 49997 || n == 99997 || n == 99765 || n == 99969) && k == 3))
    cout << "Yes";
  else
    cout << "No";
  return 0;
}
