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 N = 1e5 + 10;
int to[N], in[N], cnt[N];
int n, k, v, u;
set<int> s[N];
int main() {
  cin >> n >> k;
  if (n == 1) {
    cout << "NO" << endl;
    return 0;
  }
  for (int i = 1; i <= n - 1; i++) {
    scanf("%d%d", &v, &u);
    in[v]++;
    in[u]++;
    s[v].insert(u);
    s[u].insert(v);
  }
  int f = 0, num = 0;
  while (k--) {
    vector<int> v1;
    set<int> v2;
    for (int i = 1; i <= n; i++) {
      if (in[i] == 1) {
        v1.push_back(i);
        int tp = *(s[i].begin());
        if (in[tp] < 3) {
          f = 1;
          break;
        }
        v2.insert(tp);
        cnt[tp]++;
      } else {
        if (in[i] == 2) {
          f = 1;
          break;
        }
      }
    }
    for (int i = 0; i < v1.size(); i++) {
      int tp = *(s[v1[i]].begin());
      if (in[v1[i]] != 1 || cnt[tp] < 3) {
        f = 1;
        break;
      }
      in[tp]--;
      in[v1[i]] = 0;
      s[tp].erase(v1[i]);
      num++;
    }
    for (auto it = v2.begin(); it != v2.end(); it++) {
      if (in[*it] != 1 && k != 0) {
        f = 1;
        break;
      }
    }
    if (v2.size() <= 1) {
      break;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (in[i] != 0) f = 1;
  }
  if (num + 1 != n || k != 0) {
    cout << "NO" << endl;
    return 0;
  }
  if (f == 0) {
    cout << "YES" << endl;
  } else {
    cout << "NO" << endl;
  }
}
