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 n, k, l, r;
vector<vector<int> > v;
queue<int> q;
vector<int> dis, vis, par;
bool flag = 1;
void bfs() {
  while (!q.empty()) {
    int p = q.front();
    q.pop();
    for (int i = 0; i < (long long)(v[p].size()); i++) {
      int node = v[p][i];
      if (!vis[node]) {
        vis[node] = 1, par[node] = p;
        dis[node] = dis[p] + 1;
        q.push(node);
      }
    }
  }
}
void dfs(int s, int par, int level) {
  for (int i = 0; i < (long long)(v[s].size()); i++) {
    int node = v[s][i];
    if (node != par) {
      if ((long long)(v[node].size()) < 4 && level < k) {
        flag = 0;
      }
      dfs(node, s, level + 1);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  v.resize(n + 1);
  dis.resize(n + 1, 0);
  par.resize(n + 1, 0);
  vis.resize(n + 1, 0);
  for (int i = 0; i < n - 1; i++) {
    cin >> l >> r;
    v[l].push_back(r);
    v[r].push_back(l);
  }
  q.push(1);
  vis[1] = 1;
  bfs();
  int n1, n2, mx = -1;
  for (int i = 1; i < n + 1; i++) {
    if (dis[i] > mx) n1 = i, mx = dis[i];
    dis[i] = 0, vis[i] = 0, par[i] = 0;
  }
  q.push(n1);
  vis[n1] = 1;
  par[n1] = n1;
  bfs();
  mx = -1;
  for (int i = 1; i < n + 1; i++) {
    if (dis[i] > mx) n2 = i, mx = dis[i];
  }
  if (mx % 2 == 1)
    cout << "No\n";
  else if (mx != 2 * k)
    cout << "No\n";
  else {
    int len = mx / 2, c;
    for (int i = 0; i < len; i++) n2 = par[n2];
    c = n2;
    if ((long long)(v[c].size()) < 3) {
      cout << "No\n";
      return 0;
    }
    dfs(c, 0, 1);
    if (flag)
      cout << "Yes\n";
    else
      cout << "No\n";
  }
}
