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;
template <class S, class T>
ostream& operator<<(ostream& o, const pair<S, T>& p) {
  return o << "(" << p.first << "," << p.second << ")";
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& vc) {
  o << "{";
  for (const T& v : vc) o << v << ",";
  o << "}";
  return o;
}
using ll = long long;
template <class T>
using V = vector<T>;
template <class T>
using VV = vector<vector<T>>;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N, K;
  cin >> N >> K;
  VV<int> G(N);
  for (int i = 0; i < (int)(N - 1); i++) {
    int x, y;
    cin >> x >> y;
    x--, y--;
    G[x].push_back(y);
    G[y].push_back(x);
  }
  auto ok = [&]() {
    cout << "Yes" << endl;
    exit(0);
    return 0;
  };
  auto ng = [&]() {
    cout << "No" << endl;
    exit(0);
    return 0;
  };
  if (N <= 3) ng();
  queue<int> leaves;
  V<int> deg(N);
  for (int i = 0; i < (int)(N); i++) {
    deg[i] = G[i].size();
    if (deg[i] == 1) leaves.push(i);
  }
  V<bool> survive(N, true);
  for (int k = 0; k < (int)(K); k++) {
    leaves.push(-1);
    map<int, int> cnt;
    bool update = false;
    while (true) {
      int v = leaves.front();
      leaves.pop();
      if (v == -1) break;
      survive[v] = false;
      update = true;
      for (int u : G[v])
        if (survive[u]) {
          deg[u]--;
          cnt[u]++;
          if (deg[u] == 1) leaves.push(u);
        }
    }
    for (auto p : cnt) {
      if (p.second < 3) ng();
    }
    if (!update) ng();
  }
  int cnt = 0;
  for (int i = 0; i < (int)(N); i++)
    if (survive[i]) cnt++;
  if (cnt == 1) ok();
  ng();
}
