#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<vector<int>> g(n);
  vector<int> cur;
  for (int i = 0; i + 1 < n; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 0; i < n; i++) {
    if ((int)g[i].size() == 1) cur.push_back(i);
  }
  vector<int> vis(n);
  int depth = 1;
  while (cur.size()) {
    vector<int> nxt;
    for (int var : cur) {
      vis[var] = depth;
      for (int nbr : g[var]) {
        if (!vis[nbr]) {
          nxt.push_back(nbr);
        } else {
          if (vis[nbr] != depth - 1) {
            cout << "No" << endl;
            return 0;
          }
        }
      }
    }
    sort(nxt.begin(), nxt.end());
    nxt.erase(unique(nxt.begin(), nxt.end()), nxt.end());
    for (int var : nxt) {
      int cnt = 0;
      for (int nbr : g[var]) {
        if (vis[nbr]) cnt++;
      }
      if (cnt < 3) {
        cout << "No" << endl;
        return 0;
      }
    }
    if ((int)nxt.size() == 1) {
      cout << (depth == k ? "Yes" : "No") << endl;
      return 0;
    }
    depth++;
    swap(cur, nxt);
  }
  cout << "No" << endl;
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<vector<int>> g(n);
  vector<int> cur;
  for (int i = 0; i + 1 < n; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 0; i < n; i++) {
    if ((int)g[i].size() == 1) cur.push_back(i);
  }
  vector<int> vis(n);
  int depth = 1;
  while (cur.size()) {
    vector<int> nxt;
    for (int var : cur) {
      vis[var] = depth;
      for (int nbr : g[var]) {
        if (!vis[nbr]) {
          nxt.push_back(nbr);
        } else {
          if (vis[nbr] != depth - 1) {
            cout << "No" << endl;
            return 0;
          }
        }
      }
    }
    sort(nxt.begin(), nxt.end());
    nxt.erase(unique(nxt.begin(), nxt.end()), nxt.end());
    for (int var : nxt) {
      int cnt = 0;
      for (int nbr : g[var]) {
        if (vis[nbr]) cnt++;
      }
      if (cnt < 3) {
        cout << "No" << endl;
        return 0;
      }
    }
    if ((int)nxt.size() == 1) {
      cout << (depth == k ? "Yes" : "No") << endl;
      return 0;
    }
    depth++;
    swap(cur, nxt);
  }
  cout << "No" << endl;
  return 0;
}

```