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;
vector<int> path[100010];
int kdfs(int r, int v) {
  vector<int> lev;
  for (int sv : path[v]) {
    if (sv == r) continue;
    lev.push_back(kdfs(v, sv));
  }
  if (lev.empty()) return 0;
  if (lev.size() < 3) return -1;
  sort(lev.begin(), lev.end());
  lev.erase(unique(lev.begin(), lev.end()), lev.end());
  if (lev.size() > 1 || lev[0] < 0) {
    return -1;
  } else {
    return lev[0] + 1;
  }
}
int main() {
  int n, k;
  cin >> n >> k;
  if (n == 1) {
    cout << "No" << endl;
    return 0;
  }
  int in[n];
  fill(in, in + n, 0);
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    ++in[u], ++in[v];
    path[u].push_back(v);
    path[v].push_back(u);
  }
  queue<int> que;
  for (int v = 0; v < n; ++v) {
    if (in[v] == 1) que.push(v);
  }
  int center = -1;
  while (!que.empty()) {
    int v = que.front();
    que.pop();
    center = v;
    --in[v];
    for (int sv : path[v]) {
      if (in[sv] == 0) continue;
      --in[sv];
      if (in[sv] == 1) que.push(sv);
    }
  }
  cout << (kdfs(-1, center) == k ? "Yes" : "No") << endl;
  return 0;
}
