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;
int N, K;
vector<int> E[101010];
int D[101010];
vector<int> Ds[101010];
void solve() {
  int i, j, k, l, r, x, y;
  string s;
  cin >> N >> K;
  for (i = 0; i < (N - 1); i++) {
    cin >> x >> y;
    E[x - 1].push_back(y - 1);
    E[y - 1].push_back(x - 1);
  }
  if (N == 1) return (void)printf("No\n");
  queue<int> Q;
  for (i = 0; i < (N); i++) {
    D[i] = 101010;
    if (E[i].size() == 1) {
      D[i] = 0;
      Q.push(i);
    }
  }
  while (Q.size()) {
    x = Q.front();
    Q.pop();
    Ds[D[x]].push_back(x);
    for (auto& e : E[x])
      if (D[e] > D[x] + 1) {
        D[e] = D[x] + 1;
        Q.push(e);
      }
  }
  if (K > 100000 || Ds[K].size() != 1) return (void)printf("No\n");
  x = 0;
  for (i = K + 1; i <= 101000; i++) x += Ds[i].size();
  if (x) return (void)printf("No\n");
  for (i = 0; i < (N); i++)
    if (D[i]) {
      int num = 0;
      for (auto& e : E[i]) {
        if (D[e] == D[i] + 1)
          continue;
        else if (D[e] == D[i] - 1)
          num++;
        else
          return (void)printf("No\n");
      }
      if (num < 3) return (void)printf("No\n");
    }
  (void)printf("Yes\n");
}
int main(int argc, char** argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
  for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
  cout.tie(0);
  solve();
  return 0;
}
