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 = 1e5 + 1;
int n, k, u, v, dist[MAXN], semi, cen;
vector<int> adj[MAXN];
queue<int> q;
void dfs(int u, int d = 0, int p = -1) {
  if (u != cen && (adj[u].size() == 2 || adj[u].size() == 3)) {
    cout << "No";
    exit(0);
  }
  if (d > semi) {
    cout << "No";
    exit(0);
  }
  for (auto v : adj[u])
    if (v != p) dfs(v, d + 1, u);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) dist[i] = -1;
  for (int i = 1; i <= n - 1; ++i) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (int i = 1; i <= n; ++i)
    if (adj[i].size() == 1) {
      q.push(i);
      dist[i] = 0;
    }
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    if (semi < dist[u] && adj[u].size() > 2) {
      semi = dist[u];
      cen = u;
    }
    for (auto v : adj[u])
      if (dist[v] == -1) {
        dist[v] = dist[u] + 1;
        q.push(v);
      }
  }
  if (cen == 0 || semi != k) {
    cout << "No";
    return 0;
  }
  dfs(cen);
  cout << "Yes" << endl;
}
