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;
const int N = (int)1e5 + 10;
vector<int> tab[N], ls, cls;
int ans, dead[N], ile[N], cnt[N];
set<int> touched;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k, x, y, w;
  cin >> n >> k;
  for (int a = 1; a < n; a++) {
    cin >> x >> y;
    tab[x].push_back(y);
    tab[y].push_back(x);
  }
  for (int a = 1; a <= n; a++) {
    ile[a] = (int)tab[a].size();
    if (ile[a] == 1) ls.push_back(a);
  }
  while (!ls.empty()) {
    while (!ls.empty()) {
      w = ls.back();
      ls.pop_back();
      for (int v : tab[w]) {
        if (!dead[v]) {
          cnt[v]++;
          ile[v]--;
          touched.insert(v);
          if (ile[v] == 1) cls.push_back(v);
        }
      }
      dead[w] = 1;
    }
    while (!cls.empty()) {
      w = cls.back();
      touched.erase(w);
      cls.pop_back();
      if (cnt[w] < 3) {
        cout << "No" << endl;
        return 0;
      }
      ls.push_back(w);
    }
    if (!touched.empty()) {
      cout << "No" << endl;
      return 0;
    }
    ans++;
  }
  if (k == ans - 1) {
    cout << "Yes" << endl;
  } else
    cout << "No" << endl;
  return 0;
}
