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;
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
const double PI = acos(-1.0);
const long long INF = 1e9 + 47;
const long long LINF = INF * INF;
void bad() {
  cout << "No" << endl;
  exit(0);
}
const int MAX = 1 << 17;
int n, k;
set<int> g[MAX];
int d[MAX];
int par[MAX];
vector<int> tut;
void dfs(int v, int p = -1) {
  tut.push_back(v);
  for (auto i : g[v])
    if (i != p) par[i] = v, d[i] = d[v] + 1, dfs(i, v);
}
void check(int v, int level) {
  tut.clear();
  d[v] = 0;
  dfs(v);
  if (level == 1) {
    int cnt = 0;
    if ((int)((tut).size()) < 4) bad();
    for (auto i : tut)
      if ((int)((g[i]).size()) > 1) ++cnt;
    if (cnt > 1) bad();
    return;
  }
  if ((int)((tut).size()) < 4) bad();
  int leaf = -1;
  for (auto i : tut)
    if ((int)((g[i]).size()) == 1) {
      leaf = i;
      break;
    }
  tut.clear();
  d[leaf] = 0;
  dfs(leaf);
  int ins = -1;
  for (auto i : tut)
    if (ins == -1 || d[ins] < d[i]) ins = i;
  if (d[ins] != level * 2) bad();
  v = ins;
  for (int i = (0); i < (level); ++i) v = par[v];
  if ((int)((g[v]).size()) < 3) bad();
  for (auto i : g[v]) {
    g[i].erase(v);
    check(i, level - 1);
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  int u, v;
  for (int i = (0); i < (n - 1); ++i) {
    cin >> u >> v;
    --u;
    --v;
    g[u].insert(v);
    g[v].insert(u);
  }
  if (k > 12) bad();
  check(0, k);
  cout << "Yes" << endl;
  cerr << "Time elapsed : " << clock() / (double)CLOCKS_PER_SEC << endl;
  return 0;
}
