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;
int n, k, i, j, x, y, d[100010], mn, t, mx;
vector<vector<int> > g;
vector<int> pt;
void dfs(int v, int p, int l) {
  d[v] = l;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (to != p) {
      dfs(to, v, l + 1);
    }
  }
}
int dfs1(int v, int p, int l, int x) {
  pt.push_back(v);
  if (l == x) return 1;
  int res = 0;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (to != p) {
      res = dfs1(to, v, l + 1, x);
      if (res) return 1;
    }
  }
  pt.pop_back();
  return 0;
}
void dfs2(int v, int p, int l) {
  if (g[v].size() == 1) {
    if (l != k) {
      cout << "No\n";
      exit(0);
    }
    return;
  }
  if (l > k || g[v].size() - (p != -1) < 3) {
    cout << "No\n";
    exit(0);
  }
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (to != p) {
      dfs2(to, v, l + 1);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cin >> n >> k;
  g.resize(n + 1);
  for (i = 1; i < n; i++) {
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  if (k == 1) {
    mn = 100010;
    for (i = 1; i <= n; i++) {
      if (g[i].size() != 1) t++, mn = min(mn, (int)g[i].size());
    }
    if (t != 1 || mn < 3)
      cout << "No\n";
    else
      cout << "Yes\n";
    return 0;
  }
  dfs(1, -1, 0);
  mx = 1;
  for (i = 2; i <= n; i++) {
    if (d[i] > d[mx]) mx = i;
  }
  dfs(mx, -1, 0);
  t = mx;
  mx = 0;
  for (i = 1; i <= n; i++) mx = max(mx, d[i]);
  dfs1(t, -1, 0, mx);
  dfs2(pt[mx / 2], -1, 0);
  cout << "Yes\n";
  return 0;
}
