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 int MAXN = 100100;
int n, k, center, niv[MAXN], p1, p2, nivpfolha;
vector<int> graph[MAXN], atp, path;
bool process[MAXN], pro[MAXN];
set<int> folha;
void check(int x) {
  queue<int> q;
  q.push(center);
  memset(process, false, sizeof process);
  process[center] = 1;
  while (!q.empty()) {
    int at = q.front();
    process[at] = 1;
    q.pop();
    int grau = 0;
    for (int i = 0; i < graph[at].size(); i++) {
      int v = graph[at][i];
      niv[v] = niv[at] + 1;
      if (niv[v] > x / 2) cout << "NO", exit(0);
      if (process[v]) continue;
      if (folha.find(v) == folha.end())
        q.push(v);
      else if (niv[v] != nivpfolha)
        cout << "NO", exit(0);
      grau++;
    }
    if (grau < 3) cout << "NO", exit(0);
  }
  cout << "YES";
}
void ponta1(int x) {
  pro[x] = true;
  for (auto u : graph[x]) {
    if (pro[u]) continue;
    niv[u] = niv[x] + 1;
    if (niv[u] > niv[p1]) p1 = u;
    ponta1(u);
  }
  p1 = p1 == 0 ? x : p1;
}
void ponta2(int x) {
  pro[x] = true;
  for (auto u : graph[x]) {
    if (pro[u]) continue;
    niv[u] = niv[x] + 1;
    if (niv[u] > niv[p2]) p2 = u;
    ponta2(u);
  }
  p2 = p2 == 0 ? x : p2;
}
void caminho(int v, int t, int p) {
  if (path.size() > 0) return;
  atp.push_back(v);
  if (v == t) path = atp;
  for (auto u : graph[v])
    if (u != p) caminho(u, t, v);
  atp.pop_back();
}
vector<int> diametro(int x) {
  memset(pro, false, sizeof pro);
  p1 = p2 = 0;
  ponta1(x);
  memset(pro, false, sizeof pro);
  ponta2(p1);
  vector<int> r;
  path.clear();
  caminho(p1, p2, -1);
  r = path;
  return r;
}
int main() {
  cin >> n >> k;
  bool ok = false;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  for (int i = 1; i <= n; i++)
    if (graph[i].size() == 1) folha.insert(i);
  vector<int> diam = diametro(1);
  center = diam[((diam.size() - 1) / 2)];
  niv[center] = 0;
  nivpfolha = diam.size() / 2;
  if (diam.size() != 2 * k + 1 || diam.size() % 2 == 0) cout << "NO", exit(0);
  check(diam.size());
}
