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;
long long n, k, passed[100005], h[100005], mx, pnt, par[100005];
vector<long long> dsk[100005];
void input() {
  cin >> n >> k;
  for (long long i = 1; i < n; i++) {
    long long u, v;
    cin >> u >> v;
    dsk[u].push_back(v);
    dsk[v].push_back(u);
  }
}
void dfs(long long u) {
  passed[u] = 1;
  for (long long i = 0; i < dsk[u].size(); i++) {
    long long v = dsk[u][i];
    if (passed[v]) continue;
    par[v] = u;
    h[v] = h[u] + 1;
    dfs(v);
  }
}
void solve() {
  dfs(1);
  for (long long i = 1; i <= n; i++) {
    if (mx < h[i]) {
      mx = h[i];
      pnt = i;
    }
  }
  for (long long i = 1; i <= n; i++) {
    h[i] = 0;
    passed[i] = 0;
  }
  dfs(pnt);
  mx = 0;
  for (long long i = 1; i <= n; i++) {
    if (mx < h[i]) {
      mx = h[i];
      pnt = i;
    }
  }
  if (mx != 2 * k) {
    cout << "No";
    return;
  }
  long long f = pnt, cnt = 0;
  while (cnt < k) {
    f = par[f];
    cnt++;
  }
  for (long long i = 1; i <= n; i++) {
    passed[i] = 0;
    h[i] = 0;
  }
  if (dsk[f].size() < 3) {
    cout << "No";
    return;
  }
  dfs(f);
  for (long long i = 1; i <= n; i++) {
    if (i != f && dsk[i].size() > 1 && dsk[i].size() < 4) {
      cout << "No";
      return;
    }
    if (dsk[i].size() == 1 && h[i] != k) {
      cout << "No";
      return;
    }
  }
  cout << "Yes";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  input();
  solve();
}
