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;
std::vector<std::set<int> > graph(1e5 + 7);
std::vector<int> deg(1e5 + 7, 0), degO(1e5 + 7, 0);
std::vector<int> visited(1e5 + 7, 0), dist(1e5 + 7, 0);
void dfs(int u, int cnt) {
  visited[u] = 1;
  dist[u] = cnt;
  for (auto x : graph[u])
    if (!visited[x]) {
      dfs(x, cnt + 1);
    }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.precision(20);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    graph[a].insert(b);
    graph[b].insert(a);
    deg[a]++;
    degO[a]++;
    deg[b]++;
    degO[b]++;
  }
  int mx = 0;
  for (int i = 0; i < n + 1; ++i) {
    if (deg[i] == 1) {
      dfs(i, 0);
      for (int j = 0; j < n + 1; ++j) mx = max(dist[j], mx);
      break;
    }
  }
  if (k > mx) {
    cout << "NO\n";
    return 0;
  }
  queue<int> q;
  for (int i = 0; i < n + 1; ++i)
    if (deg[i] == 1) q.push(i);
  int cnt = 0;
  bool fake = false;
  while (q.size() && cnt < k - 1) {
    set<int> temp;
    while (q.size()) {
      int current = q.front();
      q.pop();
      deg[current] = 0;
      for (auto x : graph[current]) {
        if (degO[x] <= 3) {
          fake = true;
        }
        deg[x]--;
        temp.insert(x);
        graph[x].erase(current);
      }
    }
    for (auto x : temp) {
      q.push(x);
      if (deg[x] != 1) fake = true;
    }
    cnt++;
  }
  int limit = 0, ones = 0;
  for (int i = 0; i < n + 1; ++i) {
    if (deg[i] >= 3) limit++;
    if (deg[i] == 1) ones++;
  }
  if (limit == 1 && ones && !fake) {
    cout << "YES\n";
  } else {
    cout << "NO\n";
  }
  return 0;
}
