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;
set<int> s[100005];
int m[100005];
set<int> t, v;
int cnt[100005];
int n, k;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  cin >> n >> k;
  int x, y;
  for (int i = 1; i < n; i++) {
    cin >> x >> y;
    s[x].insert(y);
    s[y].insert(x);
    m[x]++;
    m[y]++;
  }
  for (int i = 1; i <= n; i++) {
    if (m[i] == 1) v.insert(i);
  }
  while (n > 1) {
    t.clear();
    for (auto to : v) {
      auto it = s[to].begin();
      t.insert(*it);
      cnt[*it]++;
      s[*it].erase(to);
      n--;
    }
    for (auto w : t) {
      if (cnt[w] < 3) {
        cout << "No" << endl;
        return 0;
      }
      cnt[w] = 0;
    }
    swap(t, v);
    k--;
  }
  if (k == 0)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
  return 0;
}
