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;
vector<int> v[100005];
int vis[100005], dd[100005];
int main() {
  ios_base::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n - 1; i++) {
    int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  if (n <= 3) {
    cout << "No";
    return 0;
  }
  vector<int> v1;
  for (int i = 1; i <= n; i++)
    if (v[i].size() == 1) v1.push_back(i), vis[i] = 1;
  for (int ii = 1; ii <= k; ii++) {
    set<int> prob;
    for (__typeof(v1.begin()) it = v1.begin(); it != v1.end(); it++) {
      int at = *it;
      for (__typeof(v[at].begin()) it1 = v[at].begin(); it1 != v[at].end();
           it1++) {
        int to = *it1;
        if (vis[to] == 0) prob.insert(to), dd[to]++;
      }
    }
    if (ii == k) {
      if (prob.size() != 1) {
        cout << "No";
        return 0;
      }
      int x = *prob.begin();
      if (dd[x] != v[x].size()) {
        cout << "No";
        return 0;
      }
      if (dd[x] < 3) {
        cout << "No";
        return 0;
      }
    } else {
      v1.clear();
      for (__typeof(prob.begin()) it = prob.begin(); it != prob.end(); it++) {
        vis[*it] = 1;
        v1.push_back(*it);
        if (dd[*it] < 3) {
          cout << "No";
          return 0;
        }
        if (dd[*it] != int(v[*it].size()) - 1) {
          cout << "No";
          return 0;
        }
      }
    }
  }
  cout << "Yes";
  return 0;
}
