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>
#pragma optimize("Ofast")
using namespace std;
const int N = (int)1e5 + 10;
vector<int> g[N];
int deg[N];
int level[N];
int k;
void dfs(int v, int p = -1, int depth = 0) {
  if (g[v].size() < 3 && !depth) {
    cout << "No";
    exit(0);
  }
  if ((g[v].size() == 3 || g[v].size() == 2) && depth) {
    cout << "No";
    exit(0);
  }
  int cnt = 0;
  for (int u : g[v]) {
    if (u != p) {
      dfs(u, v, depth + 1);
      ++cnt;
    }
  }
  if (!cnt && depth != k) {
    cout << "No";
    exit(0);
  }
}
signed main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  int n, ml = -1;
  cin >> n >> k;
  if (n < 4) {
    cout << "No";
    return 0;
  }
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    --u;
    --v;
    g[u].push_back(v);
    g[v].push_back(u);
    deg[v]++;
    deg[u]++;
  }
  queue<int> Q;
  for (int i = 0; i < n; ++i) {
    if (deg[i] == 1) Q.push(i);
  }
  while (!Q.empty()) {
    int v = Q.front();
    Q.pop();
    for (int i : g[v]) {
      --deg[i];
      if (deg[i] == 1) {
        level[i] = level[v] + 1;
        Q.push(i);
        ml = max(level[i], ml);
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    if (level[i] == ml) {
      dfs(i);
    }
  }
  cout << "Yes";
  return 0;
}
