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;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int n, k;
  cin >> n >> k;
  if (n < 4) {
    cout << "No";
    return 0;
  }
  vector<vector<int>> adj(n);
  for (int i = 1, u, v; i < n; ++i) {
    cin >> u >> v;
    --u, --v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  vector<int> p(n, -1);
  int mx = -1, w;
  function<void(int, int)> dfs1 = [&](int u, int d) {
    for (auto v : adj[u])
      if (v != p[u]) {
        p[v] = u;
        dfs1(v, d + 1);
      }
    if (d > mx) mx = d, w = u;
  };
  int b = 0;
  for (; b < n; ++b)
    if (adj[b].size() == 1) break;
  dfs1(b, 0);
  int tk = k;
  while (k > 0) {
    if (w != b)
      w = p[w], --k;
    else
      break;
  }
  if (k > 0) {
    cout << "No";
    return 0;
  }
  mx = -1;
  bool ok = true;
  function<void(int, int, int)> dfs = [&](int u, int p, int d) {
    int ch = 0;
    for (auto v : adj[u])
      if (v != p) {
        ++ch;
        dfs(v, u, d + 1);
      }
    ok &= ch >= 3 || d == tk;
    mx = max(mx, d);
  };
  dfs(w, -1, 0);
  ok &= mx == tk;
  if (ok)
    cout << "Yes";
  else
    cout << "No";
  return 0;
}
