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;
long long n, k, u, v, c, t, szs, cc;
vector<long long> adj[100005];
queue<long long> q;
bool vis[100005];
long long chi1[100005], chi2[100005], level[100005];
void bfs() {
  while ((long long)q.size() > 2) {
    szs = (long long)q.size();
    c++;
    for (long long i = 0; i < szs; i++) {
      t = q.front();
      q.pop();
      level[t] = c;
      if (c > 1 && chi1[t] - chi2[t] < 3) {
        cout << "No";
        return;
      }
      for (auto it : adj[t]) {
        chi2[it]--;
        if (level[it] >= level[t]) {
          cout << "No";
          return;
        }
        if (!vis[it]) {
          cc++;
          vis[it] = 1;
          q.push(it);
        }
      }
      adj[t].clear();
    }
  }
  if ((long long)q.size() != 1 || c != k || cc != n)
    cout << "No";
  else
    cout << "Yes";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  for (long long i = 0; i < n - 1; i++) {
    cin >> u >> v;
    adj[u - 1].push_back(v - 1);
    adj[v - 1].push_back(u - 1);
  }
  memset(level, -1, sizeof(level));
  for (long long i = 0; i < n; i++) {
    chi1[i] = chi2[i] = (long long)adj[i].size();
    if (chi1[i] == 1) q.push(i), vis[i] = 1, cc++;
  }
  bfs();
  return 0;
}
