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 + 100, K = 201, mod = 998244353, sq = 550, inf = 1e9 + 100,
          lg = 20;
vector<int> v[N];
int mx, g, c = -1, n, f = 0;
bool used[N];
void dfs1(int x, int p, int h) {
  if (h > mx) {
    mx = h;
    g = x;
  }
  for (auto u : v[x]) {
    if (u != p) dfs1(u, x, h + 1);
  }
}
void dfs2(int x, int p, int h) {
  if (h == 0) {
    if (used[x]) c = x;
    used[x] = 1;
  }
  for (auto u : v[x]) {
    if (u != p) dfs2(u, x, h - 1);
  }
}
void dfs3(int x, int p, int h) {
  if (h == 0 && v[x].size() != 1) {
    cout << "No";
    exit(0);
  }
  if (h != 0) {
    if (x == c && v[x].size() < 3) {
      cout << "No";
      exit(0);
    }
    if (x != c && v[x].size() < 4) {
      cout << "No";
      exit(0);
    }
  }
  for (auto u : v[x]) {
    if (u != p) dfs3(u, x, h - 1);
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int k, a, b;
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    cin >> a >> b;
    a--, b--;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  mx = 0;
  dfs1(0, -1, 0);
  a = g;
  mx = 0;
  dfs1(a, -1, 0);
  b = g;
  dfs2(a, -1, k);
  dfs2(b, -1, k);
  if (c == -1) {
    cout << "No";
    return 0;
  }
  dfs3(c, -1, k);
  cout << "Yes";
}
