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;
int constexpr N = 100005;
int n, k;
vector<vector<int>> g(N);
int center = -1;
bool dfs(int u, int p = -1, int l = 0) {
  if (l == 2 * k && g[u].size() == 1) return true;
  bool res = false;
  for (int v : g[u])
    if (v != p) {
      res |= dfs(v, u, l + 1);
      if (res && l == k) center = u;
    }
  return res;
}
bool ok = true;
void dfs2(int u, int p = -1, int l = 0) {
  if (l > k) {
    ok = false;
  }
  if (u == center) {
    ok &= (g[u].size() >= 3);
  }
  if (u != center) {
    if (l == k)
      ok &= (g[u].size() == 1);
    else
      ok &= (g[u].size() >= 4);
  }
  for (int v : g[u])
    if (v != p) {
      dfs2(v, u, l + 1);
    }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0, u, v; i < n - 1; ++i) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 1; i <= n; ++i)
    if (g[i].size() == 1) {
      dfs(i);
      break;
    }
  if (center == -1) cout << "No" << endl, exit(0);
  dfs2(center);
  cout << (ok ? "Yes" : "No") << endl;
}
