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;
using vi = vector<int>;
using vll = vector<long long int>;
template <class T>
int sgn(T x) {
  return (x > 0) - (x < 0);
}
template <typename T>
inline void read(vector<T>& v) {
  for (T& i : v) cin >> i;
}
bool solve(int caseN) {
  int n, k;
  cin >> n >> k;
  vector<vi> adj(n);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u - 1].push_back(v - 1);
    adj[v - 1].push_back(u - 1);
  }
  set<int> q;
  vector<bool> removed(n);
  for (int i = 0; i < n; i++) {
    if ((int)adj[i].size() == 1) {
      removed[i] = true;
      q.insert(i);
    }
  }
  vector<int> hasRem(n);
  while ((int)q.size() > 1) {
    k--;
    set<int> next;
    for (int i : q) {
      for (int j : adj[i]) {
        if (!removed[j]) {
          hasRem[j]++;
          next.insert(j);
        }
      }
    }
    for (int i : next) {
      if (hasRem[i] < 3 || (int)adj[i].size() - hasRem[i] >= 2) {
        cout << "No\n";
        return false;
      }
      removed[i] = true;
    }
    q = next;
  }
  if ((int)q.size() == 0) {
    cout << "No\n";
    return false;
  }
  if (k == 0) {
    cout << "Yes\n";
  } else {
    cout << "No\n";
  }
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(20);
  cout << fixed;
  for (int i = 1; solve(i); i++)
    ;
  cout.flush();
  return 0;
}
