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 + 13;
vector<int> g[N];
int pw[N];
int h[N];
int in_q[N];
bool dfs(int we, int k, int p = -1) {
  if (k != h[we]) return 0;
  int ws = 0;
  for (auto el : g[we]) {
    if (el == p) continue;
    ws++;
    if (!dfs(el, k - 1, we)) return 0;
  }
  if (k != 0 && ws < 3) {
    return 0;
  }
  return 1;
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  if (k >= n || n <= 3) {
    cout << "No" << endl;
    return 0;
  }
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  queue<int> q;
  for (int i = 1; i <= n; i++) {
    pw[i] = g[i].size();
    if (pw[i] == 1) {
      q.push(i);
      in_q[i] = 1;
    }
  }
  while (q.size() != 0) {
    int we = q.front();
    q.pop();
    for (auto el : g[we]) {
      if (in_q[el]) continue;
      pw[el]--;
      if (pw[el] == 1) {
        in_q[el] = 1;
        h[el] = h[we] + 1;
        q.push(el);
      }
    }
  }
  vector<int> cand;
  for (int i = 1; i <= n; i++) {
    if (h[i] == k) {
      cand.push_back(i);
    }
  }
  if (cand.size() != 1) {
    cout << "No\n";
    return 0;
  }
  bool res = dfs(cand[0], k);
  if (res) {
    cout << "Yes\n";
  } else {
    cout << "No\n";
  }
  return 0;
}
