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 N = 1e5 + 5;
vector<int> v[N];
int parent[N], visited[N];
vector<int> q;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int uu, vv;
    cin >> uu >> vv;
    v[uu].push_back(vv);
    v[vv].push_back(uu);
  }
  for (int i = 1; i <= n; i++) {
    if (v[i].size() == 1) {
      parent[i] = v[i][0];
      q.push_back(parent[i]);
      visited[i] = 1;
    }
  }
  sort(q.begin(), q.end());
  int ppp = 0;
  q.erase(unique(q.begin(), q.end()), q.end());
  vector<int> temp;
  int kk = 1;
  int last = 0;
  int three = 0, el = -1, xxx;
  while (q.size()) {
    int x = q.back();
    q.pop_back();
    int f = 0;
    if (v[x].size() < 3) {
      cout << "No";
      return 0;
    }
    if (v[x].size() == 3 && !visited[x]) {
      three++;
      el = x;
    }
    if (!visited[x]) {
      last = x;
      visited[x] = 1;
      for (int i : v[x]) {
        if (!visited[i]) {
          f++;
          temp.push_back(i);
        }
      }
    }
    if (f > 1) {
      cout << "No";
      return 0;
    }
    if (f == 0) {
      ppp++;
    }
    if (q.size() == 0) {
      kk++;
      q = temp;
      if (q.size() != 0) {
        sort(q.begin(), q.end());
        q.erase(unique(q.begin(), q.end()), q.end());
        temp.clear();
      }
    }
  }
  if (three > 1 || (three == 1 && el != last)) {
    cout << "No";
    return 0;
  }
  if (kk == k + 1 && ppp == 1)
    cout << "Yes";
  else
    cout << "No";
}
