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 = 100007;
int lvl[MAXN], parent[MAXN];
vector<int> edg[MAXN];
pair<int, int> dfs(int u, int p = 0) {
  lvl[u] = lvl[p] + 1;
  parent[u] = p;
  pair<int, int> rt(lvl[u], u);
  for (int v : edg[u]) {
    if (v == p) continue;
    rt = max(rt, dfs(v, u));
  }
  return rt;
}
bool solve(int k, int u, int p = 0) {
  bool flg = true;
  int cnt = 0;
  for (int v : edg[u]) {
    if (v == p) continue;
    cnt++;
    if (!solve(k - 1, v, u)) {
      flg = false;
      break;
    }
  }
  if (cnt == 0) {
    if (k == 0) return true;
    return false;
  }
  if (k == 0) return false;
  if (cnt < 3) return false;
  return flg;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    edg[u].push_back(v);
    edg[v].push_back(u);
  }
  pair<int, int> p1 = dfs(1);
  pair<int, int> p2 = dfs(p1.second);
  if (p2.first % 2 == 0) {
    cout << "No\n";
    return 0;
  }
  int root = p2.second;
  int up = p2.first / 2;
  while (up--) {
    root = parent[root];
  }
  if (solve(k, root))
    cout << "Yes\n";
  else
    cout << "No\n";
  return 0;
}
