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 n, k;
const int nax = 1e5 + 5;
const int INF = 1e9 + 7;
vector<int> adj[nax];
vector<int> dist;
vector<int> parent;
int bfs(int s) {
  dist.assign(n + 1, INF);
  parent.assign(n + 1, 0);
  dist[s] = 0;
  queue<int> q;
  q.push(s);
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    for (int w : adj[v]) {
      if (dist[w] == INF) {
        dist[w] = dist[v] + 1;
        q.push(w);
        parent[w] = v;
      }
    }
  }
  int mxV;
  int mxD = 0;
  for (int i = 1; i <= n; i++) {
    if (dist[i] > mxD) {
      mxD = dist[i];
      mxV = i;
    }
  }
  return mxV;
}
bool solve(int u, int k, int p) {
  int len = adj[u].size();
  if (p) len--;
  if (len < 3 && k > 0) return false;
  if (k == 0 && len > 0) return false;
  for (int v : adj[u]) {
    bool ret = true;
    if (v != p) ret = solve(v, k - 1, u);
    if (!ret) return false;
  }
  return true;
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  if (n <= 3) {
    cout << "No" << '\n';
    return 0;
  }
  int v = bfs(bfs(1));
  int diameter = dist[v] + 1;
  if (diameter % 2 == 0) {
    cout << "No" << '\n';
    return 0;
  }
  int center;
  int distCenter = (diameter + 1) / 2;
  while (dist[v] + 1 != distCenter) {
    v = parent[v];
  }
  center = v;
  bool is = solve(center, k, 0);
  if (is)
    cout << "Yes" << '\n';
  else
    cout << "No" << '\n';
  return 0;
}
