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;
vector<int> a[100009];
int dis[100009], par[100009];
void dfs(int x, int p) {
  dis[x] = 1 + dis[p];
  par[x] = p;
  for (int i = 0; i < a[x].size(); i++)
    if (a[x][i] != p) dfs(a[x][i], x);
}
int fun(int x, int rm, int p) {
  if (rm == 0) {
    if (a[x].size() == 1)
      return 1;
    else
      return 0;
  }
  if (a[x].size() == 1) {
    if (rm == 0) return 1;
    return 0;
  }
  int cc = 0;
  int cn = 1;
  for (int i = 0; i < a[x].size(); i++)
    if (a[x][i] != p) {
      cc++;
      cn = cn & fun(a[x][i], rm - 1, x);
    }
  if (cc > 0 && cc < 3) return 0;
  return cn;
}
void solve() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  dfs(1, 0);
  int ind = 0, mx = 0;
  for (int i = 1; i <= n; i++) {
    if (dis[i] > mx) {
      mx = dis[i];
      ind = i;
    }
    dis[i] = 0;
  }
  dfs(ind, 0);
  int node = 0;
  mx = 0;
  for (int i = 1; i <= n; i++) {
    if (dis[i] > mx) {
      mx = dis[i];
      node = i;
    }
    dis[i] = 0;
  }
  if (mx != (2 * k + 1) || n <= 3) {
    cout << "No";
    return;
  }
  int num = mx / 2;
  while (num) {
    num--;
    node = par[node];
  }
  int root = node;
  if (fun(root, k, 0))
    cout << "Yes";
  else
    cout << "No";
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
