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 maxn = 1e5 + 10;
int n, k, u, v;
int dgr[maxn], in[maxn];
set<int> se[maxn], temp;
vector<int> vc;
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &u, &v);
    se[u].insert(v);
    se[v].insert(u);
    dgr[u]++;
    dgr[v]++;
    in[u]++;
    in[v]++;
  }
  if (k > 12) return puts("No"), 0;
  for (int i = 1; i <= n; i++)
    if (dgr[i] == 1) temp.insert(i);
  for (int i = 1; i <= k; i++) {
    vc.clear();
    if (temp.empty()) return puts("No"), 0;
    for (auto it : temp) {
      vc.push_back(it);
      if (dgr[it] != 1) return puts("No"), 0;
    }
    temp.clear();
    for (auto u : vc) {
      if (se[u].size())
        v = *se[u].begin();
      else
        return puts("No"), 0;
      dgr[v]--;
      temp.insert(v);
      se[v].erase(se[v].find(u));
    }
  }
  if (temp.size() != 1 || dgr[*temp.begin()] || in[*temp.begin()] < 3)
    return puts("No"), 0;
  for (int i = 1; i <= n; i++) {
    if (in[i] == 1 || i == *temp.begin() || in[i] > 3) continue;
    return puts("No"), 0;
  }
  puts("Yes");
  return 0;
}
