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 len = 1e5 + 10;
const int mod = 998244353;
int num[len];
int in[len];
int d[len];
vector<int> ve[len];
vector<int> g;
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    ve[u].push_back(v);
    ve[v].push_back(u);
    in[u]++;
    in[v]++;
  }
  int o = 1, sum = 0;
  for (int i = 1; i <= n; ++i)
    if (in[i] == 1) g.push_back(i), num[i] = 3, d[i] = 1;
  int l = 0;
  while (k--) {
    int r = g.size();
    for (int i = l; i < r; ++i) {
      int u = g[i];
      if (num[u] < 3) return puts("No"), 0;
      num[u] = 0;
      for (int j = 0; j < ve[u].size(); ++j) {
        int v = ve[u][j];
        num[v]++;
        if (d[v] == 0)
          d[v] = d[u] + 1, g.push_back(v);
        else if (d[v] + 1 == d[u] || d[v] == d[u] + 1)
          ;
        else
          return puts("No"), 0;
      }
    }
    if (r == g.size()) return puts("No"), 0;
    if (k == 0 && g.size() == r + 1 && num[g.back()] >= 3)
      return puts("Yes"), 0;
    l = r;
  }
  puts("No");
}
