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;
vector<int> FindCenter(vector<vector<int>>& G) {
  int c = 1;
  vector<int> P;
  queue<int> q;
  for (int j = 0; j < 2; j++) {
    P.assign(G.size(), -1);
    P[c] = c;
    q.push(c);
    while (!q.empty()) {
      c = q.front();
      q.pop();
      for (auto& u : G[c])
        if (P[u] == -1) q.push(u), P[u] = c;
    }
  }
  vector<int> diameter = {c};
  while (P[c] != c) c = P[c], diameter.push_back(c);
  if (diameter.size() % 2 == 1) return {diameter[diameter.size() / 2]};
  return {diameter[diameter.size() / 2 - 1], diameter[diameter.size() / 2]};
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<vector<int>> G(n + 1);
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  int center = FindCenter(G)[0];
  queue<int> q;
  vector<int> D(n + 1, INT_MAX);
  D[center] = 0;
  q.push(center);
  if (G[center].size() < 3) return cout << "No", 0;
  while (!q.empty()) {
    int c = q.front();
    q.pop();
    if (G[c].size() == 1 && D[c] != k) return cout << "No", 0;
    if (G[c].size() < 4 && D[c] < k && c != center) return cout << "No", 0;
    for (auto& u : G[c])
      if (D[u] == INT_MAX) D[u] = D[c] + 1, q.push(u);
  }
  cout << "Yes\n";
}
