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 long long inf = 1e18;
const long long maxu = 100005;
vector<long long> adj[maxu];
vector<long long> vis(maxu);
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long i, j, k = 0, n, t = 1, m, l = 0;
  while (t--) {
    cin >> n >> k;
    for (i = 0; i <= n; i++) vis[i] = 0;
    for (i = 0; i <= n; i++) {
      adj[i].clear();
    }
    for (i = 0; i < n - 1; i++) {
      cin >> l >> m;
      adj[l].push_back(m);
      adj[m].push_back(l);
    }
    if (n <= 3) {
      cout << "No\n";
      continue;
    }
    long long d[n + 1];
    bool ans = true;
    for (i = 0; i <= n; i++) d[i] = 100000000000;
    queue<long long> q;
    for (i = 1; i <= n; i++) {
      sort(adj[i].begin(), adj[i].end());
      if (adj[i].size() == 1) {
        d[i] = 0;
        q.push(i);
        vis[i] = 1;
      }
    }
    long long cnt = 0;
    long long maxi = -1;
    long long leaf[n + 1];
    for (i = 1; i <= n; i++) {
      leaf[i] = 0;
    }
    while (!q.empty()) {
      long long s = q.front();
      q.pop();
      if (adj[s].size() > 1) {
        ans = false;
        break;
      } else if (adj[s].size() == 0) {
        cnt++;
        if (!q.empty()) {
          ans = false;
          break;
        }
      }
      if (d[s] && leaf[s] < 3) {
        ans = false;
        break;
      }
      long long cnt1 = 0;
      for (auto u : adj[s]) {
        cnt1++;
        leaf[u]++;
        adj[u].erase(lower_bound(adj[u].begin(), adj[u].end(), s));
        if (vis[u]) continue;
        vis[u] = 1;
        d[u] = d[s] + 1;
        maxi = max(maxi, d[u]);
        q.push(u);
      }
      if (cnt1 > 1) {
        ans = false;
        break;
      } else {
        adj[s].clear();
      }
    }
    long long cnt2 = 0;
    for (i = 1; i <= n; i++) {
      if (d[i] == maxi) cnt2++;
    }
    if (cnt2 == 1 && cnt == 1 && ans && maxi == k) {
      cout << "Yes\n";
    } else
      cout << "No\n";
  }
  return 0;
}
