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;
using nagai = long long;
using ll = long long;
const int N = 100000 + 10;
vector<int> g[N];
void fail() {
  cout << "NO\n";
  exit(0);
}
int k = 0;
void d(int u, int p, int curd) {
  if (p == -1) {
    if (g[u].size() < 3) fail();
  } else {
    if (g[u].size() < 4 && g[u].size() != 1) {
      cerr << "kek\n";
      fail();
    }
    if (curd > k || (g[u].size() == 1 && curd < k)) {
      cerr << curd << ' ' << k << '\n';
      fail();
    }
  }
  for (int v : g[u]) {
    if (v == p) continue;
    d(v, u, curd + 1);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n >> k;
  if (k > 30) {
    cout << "NO\n";
    return 0;
  }
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  vector<int> q = {0};
  vector<int> dist(n, -1);
  vector<int> prev(n, -1);
  dist[0] = 0;
  for (int i = 0; i < q.size(); ++i) {
    int x = q[i];
    for (int y : g[x])
      if (dist[y] == -1) dist[y] = dist[x] + 1, q.emplace_back(y), prev[y] = x;
  }
  fill(prev.begin(), prev.end(), -1);
  int mem = max_element(dist.begin(), dist.end()) - dist.begin();
  q = {mem};
  fill(dist.begin(), dist.end(), -1);
  fill(prev.begin(), prev.end(), -1);
  dist[mem] = 0;
  for (int i = 0; i < q.size(); ++i) {
    int x = q[i];
    for (int y : g[x])
      if (dist[y] == -1) dist[y] = dist[x] + 1, q.emplace_back(y), prev[y] = x;
  }
  int kek = max_element(dist.begin(), dist.end()) - dist.begin();
  vector<int> path;
  while (kek != mem) path.push_back(kek), kek = prev[kek];
  path.push_back(mem);
  if (path.size() % 2 == 0) fail();
  int cent = path[path.size() / 2];
  d(cent, -1, 0);
  cout << "YES\n";
  return 0;
}
