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, k2;
int gr[100005], gr2[100005], in[100005];
set<int> v[100005];
vector<int> q[100005];
inline bool solve(int k) {
  if (q[k].empty()) return 0;
  if (k2 == k && q[k].size() > 1) return 0;
  if (k2 == k && v[q[k][0]].size() > 0) return 0;
  if (k2 == k) return 1;
  if (q[k].size() < 3) return 0;
  for (auto nod : q[k]) {
    for (auto it : v[nod]) {
      --gr[it];
      ++in[it];
      if (gr[it] < 0) return 0;
      v[it].erase(nod);
      if (gr[it] == 0 && k + 1 == k2)
        q[k + 1].push_back(it);
      else if (gr[it] == 0)
        return 0;
      else if (gr[it] == 1 && k + 1 != k2)
        q[k + 1].push_back(it), --gr[it];
    }
  }
  for (auto nod : q[k]) {
    for (auto it : v[nod]) {
      if (gr[it] > 1) return 0;
    }
  }
  return 1;
}
int main() {
  scanf("%d%d", &n, &k);
  k2 = k;
  int x, y;
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &x, &y);
    v[x].insert(y);
    v[y].insert(x);
  }
  for (int i = 1; i <= n; ++i) {
    gr[i] = v[i].size();
    gr2[i] = gr[i];
    if (gr[i] == 2) {
      printf("No");
      return 0;
    }
  }
  for (int i = 1; i <= n; ++i)
    if (gr[i] == 1) q[0].push_back(i), --gr[i];
  for (int i = 0; i <= k; ++i) {
    bool ok = solve(i);
    if (!ok) {
      printf("No");
      return 0;
    }
  }
  for (int i = 1; i <= n; ++i)
    if (in[i] > 0 && in[i] < 3) {
      printf("No");
      return 0;
    }
  printf("Yes");
  return 0;
}
