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 + 1;
vector<int> adj[N];
int dep[N];
int par[N];
int k;
void dfs_dep(int u, int p) {
  for (int c : adj[u]) {
    if (c != p) {
      dep[c] = dep[u] + 1;
      par[c] = u;
      dfs_dep(c, u);
    }
  }
}
void dfs_check(int u, int p) {
  if (dep[u] < k && int(adj[u].size()) + !p < 4) {
    cout << "No\n";
    exit(0);
  }
  if (dep[u] >= k && int(adj[u].size()) > 1) {
    cout << "No\n";
    exit(0);
  }
  for (int c : adj[u]) {
    if (c != p) {
      dfs_check(c, u);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n >> k;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs_dep(1, 0);
  int root = max_element(dep, dep + n + 1) - dep;
  dep[root] = 0;
  dfs_dep(root, 0);
  root = max_element(dep, dep + n + 1) - dep;
  int len = dep[root];
  for (int i = 0; i < len / 2; ++i) {
    root = par[root];
  }
  if (len != 2 * k) {
    cout << "No\n";
    exit(0);
  }
  dep[root] = 0;
  dfs_dep(root, 0);
  dfs_check(root, 0);
  cout << "Yes\n";
}
