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 = 200000 + 10;
struct Node {
  int p;
  int dis;
};
vector<pair<int, int> > tree[maxn];
int n, m, u, v, len, cen, du[maxn], dv[maxn], vis[maxn];
int flp(int s) {
  memset(vis, 0, sizeof(vis));
  int max_dis = 0, lp = -1;
  queue<Node> q;
  q.push(Node{s, 0});
  vis[s] = 1;
  while (!q.empty()) {
    Node f = q.front();
    q.pop();
    if (s == u) du[f.p] = f.dis;
    if (s == v) dv[f.p] = f.dis;
    if (f.dis > max_dis) {
      max_dis = f.dis;
      lp = f.p;
    }
    for (int i = 0; i < tree[f.p].size(); i++) {
      int next = tree[f.p][i].first;
      int d = tree[f.p][i].second;
      if (vis[next]) continue;
      vis[next] = 1;
      q.push(Node{next, f.dis + d});
    }
  }
  return lp;
}
bool dfs(int w, int fa, int dep) {
  int siz = tree[w].size();
  for (int i = 0; i < siz; i++) {
    pair<int, int> v = tree[w][i];
    if (v.first == fa) continue;
    if (dfs(v.first, w, dep + 1) == false) return false;
  }
  if (cen == w)
    return (siz >= 3);
  else if (dep < m)
    return (siz >= 4);
  else
    return (siz == 1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int a, b;
  cin >> n >> m;
  for (int i = 1; i < n; i++) {
    cin >> a >> b;
    tree[a].push_back(make_pair(b, 1));
    tree[b].push_back(make_pair(a, 1));
  }
  u = flp(1);
  v = flp(u);
  flp(v);
  len = du[v];
  cen = -1;
  for (int i = 1; i <= n; i++)
    if (dv[i] == du[i] && (dv[i] + du[i]) == len) {
      cen = i;
      break;
    }
  if (cen == -1) {
    cout << "No" << endl;
    return 0;
  }
  if (dfs(cen, -1, 0) == true) {
    cout << "Yes" << endl;
  } else
    cout << "No" << endl;
  return 0;
}
