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(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> adj[n];
  int deg[n];
  for (int i = 0; i < n; i++) {
    deg[i] = 0;
  }
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
    deg[a]++;
    deg[b]++;
  }
  vector<int> li;
  for (int i = 0; i < n; i++) {
    if (deg[i] == 1) {
      li.push_back(i);
    }
  }
  for (int i = 0; i < li.size(); i++) {
    int now = li[i];
    for (int j = 0; j < adj[now].size(); j++) {
      int to = adj[now][j];
      deg[to]--;
      if (deg[to] == 1) {
        li.push_back(to);
      }
    }
  }
  if (li.size() != n) {
    cout << "No" << endl;
    return 0;
  }
  bool v[n];
  int deep[n];
  for (int i = 0; i < n; i++) {
    v[i] = false;
    deep[i] = -1;
  }
  v[li[n - 1]] = true;
  deep[li[n - 1]] = 0;
  bool ok = true;
  for (int i = n - 1; i >= 0; i--) {
    int now = li[i];
    if (!v[now] || deep[now] > k) {
      ok = false;
      break;
    }
    int red = adj[now].size();
    if (i < n - 1) {
      red--;
    }
    if (red < 3 && deep[now] != k) {
      ok = false;
      break;
    }
    for (int j = 0; j < adj[now].size(); j++) {
      int to = adj[now][j];
      if (!v[to]) {
        v[to] = true;
        deep[to] = deep[now] + 1;
      }
    }
  }
  cout << (ok ? "Yes" : "No") << endl;
}
