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;
using namespace std::chrono;
const long long int mod = 1000000007ll;
const long long int mod2 = 998244353ll;
const long long int inf = 10000000000000007ll;
const double eps = 1e-10;
template <typename T = long long int, T D = 0>
class MaxOp {
 public:
  static T op(const T& a, const T& b) { return max(a, b); }
  static T def() { return (T)D; }
};
template <typename T = long long int, T D = inf>
class MinOp {
 public:
  static T op(const T& a, const T& b) { return min(a, b); }
  static T def() { return (T)D; }
};
template <typename T = long long int, T D = 0>
class SumOp {
 public:
  static T op(const T& a, const T& b) { return a + b; }
  static T def() { return (T)D; }
};
template <typename D, typename T = long long int>
class SegTree {
 public:
  long long int n;
  vector<T> t;

 private:
  void build(const long long int* a, int v, int tl, int tr) {
    if (tl == tr)
      t[v] = a[tl];
    else {
      int tm = (tl + tr) / 2;
      build(a, v * 2, tl, tm);
      build(a, v * 2 + 1, tm + 1, tr);
      t[v] = lca(t[v * 2], t[v * 2 + 1]);
    }
  }
  long long int get(int v, int tl, int tr, int l, int r) {
    if (l > r) return D().def();
    if (l == tl && r == tr) return t[v];
    int tm = (tl + tr) / 2;
    return D().op(get(v * 2, tl, tm, l, min(r, tm)),
                  get(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r));
  }

 public:
  void build(const vector<long long int>& a) {
    n = a.size();
    t.resize(n * 4);
    build(a.data(), 1, 0, n - 1);
  }
  void build(long long int* a, long long int n) {
    this->n = n;
    t.resize(n * 4);
    build(a, 1, 0, n - 1);
  }
  long long int get(long long int l, long long int r) {
    return get(1, 0, n - 1, l, r);
  }
};
long long int n, k;
vector<long long int> e[100005];
long long int v[100005];
map<long long int, long long int> q[100005];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout.precision(15);
  cout.setf(ios::fixed);
  cin >> n >> k;
  for (long long int i = 1; i < n; i++) {
    long long int t1, t2;
    cin >> t1 >> t2;
    e[t1].push_back(t2);
    e[t2].push_back(t1);
  }
  for (long long int i = 1; i <= n; i++) {
    if (e[i].size() == 1) {
      q[0].insert({i, 3});
    }
  }
  long long int t = 0;
  while (t < k) {
    for (auto i : q[t]) {
      v[i.first] = 1;
    }
    for (auto i : q[t]) {
      if (i.second < 3) {
        cout << "No" << endl;
        return 0;
      }
      bool f = false;
      for (auto j : e[i.first]) {
        if (v[j]) {
          continue;
        }
        f = 1;
        q[t + 1][j]++;
      }
      if (!f) {
        cout << "No" << endl;
        return 0;
      }
    }
    t++;
  }
  if (t == k && q[t].size() == 1 && q[t].begin()->second >= 3) {
    cout << "Yes" << endl;
  } else {
    cout << "No" << endl;
  }
  return 0;
}
