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;
int f[N], c[N];
set<int> v[N];
vector<int> q, q2;
int main() {
  int n, k, cnt = 0;
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    v[a].insert(b);
    v[b].insert(a);
  }
  for (int i = 1; i <= n; i++)
    if (v[i].size() == 1) q.push_back(i);
  while (1) {
    cnt++;
    if (cnt > k) {
      cout << "No\n";
      return 0;
    }
    for (int i = 0; i < q.size(); i++) {
      int x = q[i];
      if (!v[x].size()) {
        cout << "No\n";
        return 0;
      }
      int p = (*v[x].begin());
      if (v[p].find(x) == v[p].end()) {
        cout << "No\n";
        return 0;
      }
      v[p].erase(v[p].find(x)), c[p]++;
    }
    for (int i = 0; i < q.size(); i++) {
      int x = q[i];
      if (!v[x].size()) {
        cout << "No\n";
        return 0;
      }
      int p = (*v[x].begin());
      if (!f[p]) {
        f[p] = 1;
        if (v[p].size() == 1 && c[p] > 2)
          q2.push_back(p);
        else if (v[p].size() == 0 && cnt == k && c[p] > 2) {
          cout << "Yes\n";
          return 0;
        } else {
          cout << "No\n";
          return 0;
        }
      }
    }
    q.clear();
    q = q2;
    q2.clear();
  }
  cout << "Yes\n";
  return 0;
}
