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 nmax = 1e5 + 3;
int n, m, a, b, i, nivmax, jman[nmax], haida[nmax], fol[nmax];
vector<int> t2[nmax], t1[nmax], haicaca[nmax];
void usurelu(int nod, int niv);
string p1 = "Yes";
string p2 = "No";
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 1; i < n; ++i) {
    cin >> a >> b;
    t1[a].push_back(b);
    t1[b].push_back(a);
    jman[a] += 1;
    jman[b] += 1;
    haida[a] += 1;
    haida[b] += 1;
  }
  for (int i = 1; i <= n; ++i)
    if (haida[i] == 1) t2[1].push_back(i);
  for (int i = 1; i <= m; ++i) {
    if (!t2[i].size()) {
      cout << p2;
      return 0;
    }
    for (auto it : t2[i]) {
      for (auto nanu : t1[it]) {
        haida[nanu] -= 1;
        if (haida[nanu] == 1) t2[i + 1].push_back(nanu);
      }
    }
  }
  if (t2[m + 1].size() != 1) {
    cout << p2;
    return 0;
  }
  usurelu(t2[m + 1][0], 1);
  for (int i = 2; i < nivmax; ++i) {
    for (auto it : haicaca[i]) {
      if (jman[it] < 4) {
        cout << p2;
        return 0;
      }
    }
  }
  if (jman[haicaca[1][0]] < 3)
    cout << p2;
  else
    cout << p1;
  return 0;
}
void usurelu(int nod, int niv) {
  fol[nod] = 1;
  haicaca[niv].push_back(nod);
  nivmax = max(nivmax, niv);
  for (auto it : t1[nod])
    if (!fol[it]) usurelu(it, niv + 1);
}
