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;
const int N = 1e5 + 22;
int n, k;
int zz[N], zz2[N];
bool vis[N];
set<pair<int, int> > q;
vector<int> G[N];
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  int x, y;
  cin >> n >> k;
  for (int i = 0; i < (n - 1); i++) {
    cin >> x >> y;
    x--, y--;
    G[x].push_back(y);
    G[y].push_back(x);
    zz[x]++;
    zz2[x]++;
    zz[y]++;
    zz2[y]++;
  }
  for (int i = 0; i < (n); i++) {
    q.insert({int((G[i]).size()), i});
  }
  while (k) {
    vector<int> vec;
    while (!q.empty()) {
      auto it = q.begin();
      if (it->first != 1) break;
      int sw = -1;
      for (auto &u : (G[it->second])) {
        if (!vis[u]) vec.push_back(u), sw = 1, zz2[u]--;
      }
      vis[it->second] = 1;
      q.erase(it);
      if (sw == -1) return cout << "No", 0;
    }
    (sort(vec.begin(), vec.end()),
     vec.resize(distance(vec.begin(), unique(vec.begin(), vec.end()))));
    if (vec.empty()) return cout << "No", 0;
    for (auto &u : (vec)) {
      if (zz2[u] != 1 && k != 1) return cout << "No", 0;
      if (zz[u] < 4 && k != 1) return cout << "No", 0;
      q.erase({zz[u], u});
      q.insert({1, u});
    }
    k--;
  }
  if (q.size() == 1 && zz[q.begin()->second] >= 3)
    cout << "Yes";
  else
    cout << "No";
  return 0;
}
