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 dp[111111];
vector<long long> g[111111];
long long k;
bool fl = 1;
long long maxi = -1;
long long ind;
long long ind1;
long long d[111111];
long long pa[111111];
void dfs_penis(long long v, long long p, long long dpth) {
  d[v] = dpth;
  pa[v] = p;
  for (auto to : g[v]) {
    if (to == p) continue;
    dfs_penis(to, v, dpth + 1);
  }
}
void dfs(long long v, long long p, long long dpth) {
  if (g[v].size() == 1) {
    if (dpth != k) fl = 0;
  } else if (g[v].size() <= 3 && p != -1)
    fl = 0;
  for (auto to : g[v]) {
    if (to == p) continue;
    dfs(to, v, dpth + 1);
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n >> k;
  if (n == 1) {
    cout << "No";
    return 0;
  }
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    g[u - 1].push_back(v - 1);
    g[v - 1].push_back(u - 1);
  }
  long long leave = -1;
  for (long long i = 0; i < n; i++) {
    if (g[i].size() == 1) leave = i;
  }
  dfs_penis(leave, -1, 0);
  long long maxi = 0;
  long long ind = 0;
  for (long long i = 0; i < n; i++) {
    if (d[i] > maxi) {
      maxi = d[i];
      ind = i;
    }
  }
  if (maxi != 2 * k) fl = 0;
  vector<long long> path;
  for (; ind != leave; ind = pa[ind]) {
    path.push_back(ind);
  }
  path.push_back(leave);
  if (k >= path.size()) {
    cout << "No";
    return 0;
  }
  ind1 = path[k];
  if (g[ind1].size() < 3) fl = 0;
  dfs(ind1, -1, 0);
  fl == 1 ? cout << "Yes" : cout << "No";
  return 0;
}
