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 maxn = 1e5;
set<int> s[maxn + 5];
set<pair<int, int> > q;
int du[maxn + 5], cnt[maxn + 5];
int n, k, x, y, lev;
int main(void) {
  cin >> n >> k;
  for (int i = 1; i < n; ++i) {
    cin >> x >> y;
    s[x].insert(y);
    s[y].insert(x);
    ++du[x];
    ++du[y];
  }
  for (int i = 1; i <= n; ++i)
    if (s[i].size() == 1) q.insert(pair<int, int>(lev, i));
  while (!q.empty()) {
    int p = q.size();
    if (p == 1) break;
    ++lev;
    for (int i = 1; i <= p; ++i) {
      x = q.begin()->second;
      q.erase(q.begin());
      if (s[x].size() != 1) {
        cout << "No" << endl;
        return 0;
      }
      y = *s[x].begin();
      ++cnt[y];
      s[y].erase(x);
      q.insert(pair<int, int>(lev, y));
    }
  }
  if (lev != k) {
    cout << "No\n";
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    if (cnt[i] && cnt[i] < 3) {
      cout << "No\n" << endl;
      return 0;
    }
  }
  cout << "Yes" << endl;
  return 0;
}
