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;
void solve() {}
const long long maxn = 100500;
long long n, k;
vector<long long> G[maxn];
long long dis[maxn];
long long par[maxn];
void no() {
  cout << "No" << '\n';
  exit(0);
}
void cl() {
  for (long long i = 0; i < maxn; i++) dis[i] = 1e18;
}
void dfs(long long v, long long p) {
  long long ch = 0;
  long long leaves = 0;
  for (long long to : G[v]) {
    if (to == p) continue;
    ch++;
    if ((long long)G[to].size() == 1) leaves++;
    dfs(to, v);
  }
  if (ch == 0)
    return;
  else if (ch < 3)
    no();
  else if (leaves != ch and leaves != 0) {
    if (false) cerr << "failed on " << v << '\n';
    no();
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (long long i = 1; i <= n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    G[u].emplace_back(v);
    G[v].emplace_back(u);
  }
  cl();
  queue<long long> Q;
  Q.push(1);
  dis[1] = 0;
  pair<long long, long long> cur = {-1e18, -1};
  while (!Q.empty()) {
    long long S = Q.front();
    Q.pop();
    cur = max(cur, pair<long long, long long>(dis[S], S));
    for (long long to : G[S]) {
      if (dis[to] > dis[S] + 1) {
        dis[to] = dis[S] + 1;
        Q.push(to);
      }
    }
  }
  long long e = cur.second;
  if (false) cerr << "furhtest away from 1 is " << e << '\n';
  cl();
  dis[e] = 0;
  Q.push(e);
  cur = {-1e18, -1};
  while (!Q.empty()) {
    long long S = Q.front();
    Q.pop();
    cur = max(cur, pair<long long, long long>(dis[S], S));
    for (long long to : G[S]) {
      if (dis[to] > dis[S] + 1) {
        par[to] = S;
        dis[to] = dis[S] + 1;
        Q.push(to);
      }
    }
  }
  vector<long long> path;
  long long c = cur.second;
  if (false) cerr << "furthest away from " << e << " is " << c << '\n';
  while (c != e) {
    path.emplace_back(c);
    c = par[c];
  }
  path.emplace_back(e);
  for (long long x : path) {
    if (false) cerr << x << " ";
  }
  if (false) cerr << '\n';
  if ((long long)path.size() != 2 * k + 1) no();
  long long mid = path[(long long)path.size() / 2];
  dfs(mid, -1);
  cout << "Yes" << '\n';
}
