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;
set<int> adj[100005];
int visit[100005], c[100005];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    adj[a].insert(b);
    adj[b].insert(a);
  }
  set<int> cur;
  for (int i = 1; i <= n; i++) {
    if (adj[i].size() == 1) cur.insert(i);
  }
  int cnt = 0, lvl = 0;
  while (cnt < n - 1) {
    lvl++;
    set<int> tmp;
    while (cur.size() > 0) {
      cnt++;
      int xx = *(cur.begin());
      cur.erase(cur.begin());
      if (adj[xx].size() == 1) {
        int par = *(adj[xx].begin());
        tmp.insert(par);
        adj[par].erase(xx);
        c[par]++;
      } else {
        printf("No");
        return 0;
      }
    }
    for (set<int>::iterator it = tmp.begin(); it != tmp.end(); it++) {
      if (c[*it] < 3) {
        printf("No");
        return 0;
      }
      cur.insert(*it);
    }
  }
  if (lvl == k) {
    printf("Yes\n");
  } else
    printf("No\n");
}
