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;
vector<vector<int> > G, G_or;
vector<int> d;
vector<char> used;
void dfs1(int v) {
  used[v] = 1;
  for (auto to : G[v]) {
    if (used[to]) continue;
    G_or[v].push_back(to);
    d[to] = d[v] + 1;
    dfs1(to);
  }
}
int main() {
  int n, k;
  cin >> n >> k;
  used.assign(n, 0);
  d.assign(n, 0);
  G.assign(n, vector<int>(0));
  G_or.assign(n, vector<int>(0));
  queue<int> bfs;
  for (int i = 0, a, b; i < n - 1; i++) {
    cin >> a >> b;
    --a;
    --b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  for (int v = 0; v < n; v++) {
    if (G[v].size() == 1) {
      used[v] = 1;
      bfs.push(v);
    }
  }
  while (bfs.size()) {
    int v = bfs.front();
    bfs.pop();
    for (auto to : G[v]) {
      if (!used[to]) {
        bfs.push(to);
        used[to] = 1;
        d[to] = d[v] + 1;
      }
    }
  }
  int root = 0;
  for (int v = 0; v < n; v++) {
    if (d[v] > d[root]) root = v;
  }
  used.assign(n, 0);
  d.assign(n, 0);
  dfs1(root);
  for (int v = 0; v < n; v++) {
    if (d[v] == k) {
      if (G_or[v].size() != 0) return cout << "No", 0;
    } else if (G_or[v].size() < 3)
      return cout << "No", 0;
  }
  cout << "Yes";
  return 0;
}
