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;
const int64_t N = 1e5 + 1;
int64_t crt_node = 0, maxi = 0;
vector<int64_t> previous;
vector<vector<int64_t>> adj;
void dfs(int64_t node, int64_t pere, int64_t len) {
  previous[node] = pere;
  if (len > maxi) {
    crt_node = node;
    maxi = len;
  }
  for (int64_t i : adj[node]) {
    if (i != pere) dfs(i, node, len + 1);
  }
}
bool ok = true;
vector<int64_t> level, path;
void hedgehog(int64_t node, int64_t pere) {
  int64_t k = 0;
  for (int64_t i : adj[node]) {
    if (i != pere) {
      hedgehog(i, node);
      if (k == 0)
        k = level[i];
      else if (k != level[i])
        ok = false;
    }
  }
  if (adj[node].size() == 1)
    level[node] = 1;
  else if (adj[node].size() < 3 or
           (adj[node].size() == 3 and path[path.size() / 2] != node))
    ok = false;
  else
    level[node] = k + 1;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int64_t n, k;
  cin >> n >> k;
  if (n == 1) {
    cout << "No";
    return 0;
  }
  adj.resize(n + 1);
  for (int64_t i = 1; i < n; ++i) {
    int64_t u, v;
    cin >> u >> v;
    adj[u].emplace_back(v);
    adj[v].emplace_back(u);
  }
  previous.resize(n + 1);
  dfs(1, 0, 0);
  int64_t a = crt_node;
  maxi = 0;
  fill(previous.begin(), previous.end(), 0ll);
  dfs(crt_node, 0, 0);
  int64_t b = crt_node;
  while (b) {
    path.emplace_back(b);
    b = previous[b];
  }
  level.resize(n + 1);
  hedgehog(path[path.size() / 2], 0);
  if (level[path[path.size() / 2]] != k + 1) ok = false;
  cout << (ok ? "Yes" : "No");
  return 0;
}
