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<vector<int>> adj;
vector<int> vis, p, d;
bool ok;
int k, c;
void dfs(int u) {
  for (int v : adj[u]) {
    if (vis[v] == 0) {
      vis[v] = 1;
      p[v] = u;
      d[v] = d[u] + 1;
      dfs(v);
    }
  }
}
void dfs1(int u, int d) {
  if (((d < k) && adj[u].size() < 4 - (u == c)) ||
      (d == k) && adj[u].size() != 1)
    ok = 0;
  for (int v : adj[u]) {
    if (vis[v] == 0) {
      vis[v] = 1;
      dfs1(v, d + 1);
    }
  }
}
int main() {
  int n;
  cin >> n >> k;
  adj.resize(n + 1);
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  vis.resize(n + 1);
  p.resize(n + 1);
  d.resize(n + 1);
  for (int i = 1; i <= n; i++) {
    if (adj[i].size() == 1) {
      vis[i] = 1, dfs(i);
      break;
    }
  }
  int e = distance(d.begin(), max_element(d.begin(), d.end()));
  int dia = d[e];
  if (dia / 2 != k) {
    cout << "No\n";
    return 0;
  }
  while (d[e] != dia / 2) e = p[e];
  vis.assign(n + 1, 0);
  ok = 1;
  c = e;
  vis[c] = 1;
  dfs1(c, 0);
  cout << (ok ? "Yes" : "No") << '\n';
}
