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 + 7;
const int INF = 0x3f3f3f3f;
const long long mod = 998244353;
int n, k;
set<int> st[maxn];
int cnt[maxn];
int d[maxn];
int main() {
  scanf("%d%d", &n, &k);
  if (n < 4) {
    printf("No");
    return 0;
  }
  int u, v;
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &u, &v);
    st[u].insert(v);
    st[v].insert(u);
    d[u]++, d[v]++;
  }
  int num = 0;
  set<int> t;
  for (int i = 1; i <= n; ++i) {
    if (d[i] == 1) {
      int j = *st[i].begin();
      t.insert(j);
      cnt[j]++;
      st[j].erase(st[j].find(i));
      num++;
    }
  }
  int ans = 1;
  set<int> t_;
  while (t.size()) {
    if (t.size() == 1) {
      int j = *t.begin();
      if (cnt[j] < 3) {
        printf("No");
        return 0;
      }
      num++;
      break;
    }
    for (auto i : t) {
      if (cnt[i] < 3 || st[i].size() != 1) {
        printf("No");
        return 0;
      }
      int j = *st[i].begin();
      t_.insert(j);
      cnt[j]++;
      st[j].erase(st[j].find(i));
      num++;
    }
    t.clear();
    for (auto i : t_) {
      t.insert(i);
    }
    t_.clear();
    ans++;
  }
  if (ans == k)
    printf("Yes");
  else
    printf("No");
  return 0;
}
