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 = 100010;
int n, k, ind[maxn], vis[maxn];
set<int> G[maxn], s, t;
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d %d", &u, &v);
    G[u].insert(v), G[v].insert(u);
  }
  for (int i = 1; i <= n; i++) {
    if (G[i].size() == 1) s.insert(i);
  }
  for (; n > 1; swap(s, t), k--) {
    t.clear();
    for (int v : s) {
      int u = *G[v].begin();
      t.insert(u), G[u].erase(v);
      ind[u]++, n--;
    }
    for (int v : t) {
      if (ind[v] < 3) printf("No\n"), exit(0);
      ind[v] = 0;
    }
  }
  printf("%s\n", !k ? "Yes" : "No");
  return 0;
}
