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;
using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
constexpr ll TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n - 1); }
template <class T>
using V = vector<T>;
template <class T>
using VV = V<V<T>>;
VV<int> g;
int ma_v, ma_dps;
void dfs(int p, int b, int ndp = 0) {
  if (ma_dps < ndp) {
    ma_dps = ndp;
    ma_v = p;
  }
  for (int d : g[p]) {
    if (d == b) continue;
    dfs(d, p, ndp + 1);
  }
}
V<int> path, buf;
void dfs2(int p, int b, int t) {
  buf.push_back(p);
  if (p == t) path = buf;
  for (int d : g[p]) {
    if (d == b) continue;
    dfs2(d, p, t);
  }
  buf.pop_back();
}
void fu() {
  cout << "No" << endl;
  exit(0);
}
void gg() {
  cout << "Yes" << endl;
  exit(0);
}
int k;
void check(int p, int b, int ndp = 0) {
  int deg = int(g[p].size());
  if (ndp == 0) {
    if (deg < 3) fu();
  } else if (ndp < k) {
    if (deg < 4) fu();
  } else {
    if (deg != 1) fu();
  }
  for (int d : g[p]) {
    if (d == b) continue;
    check(d, p, ndp + 1);
  }
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << setprecision(20) << fixed;
  int n;
  cin >> n >> k;
  g = VV<int>(n);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  ma_dps = 0;
  dfs(0, -1);
  int u = ma_v;
  ma_dps = 0;
  dfs(u, -1);
  int v = ma_v;
  ma_dps = 0;
  dfs2(u, -1, v);
  int m = int(path.size());
  int mid = path[m / 2];
  check(mid, -1);
  gg();
  return 0;
}
