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;
long long k;
vector<int> e[100005];
int d[100005][2];
pair<int, int> far(int cn, int p) {
  pair<int, int> ans = {-1, cn};
  for (auto nn : e[cn])
    if (nn != p) {
      pair<int, int> res = far(nn, cn);
      ans = max(ans, res);
    }
  ++ans.first;
  return ans;
}
pair<int, pair<int, int> > diam() {
  pair<int, int> res1 = far(1, 0);
  pair<int, int> res2 = far(res1.second, 0);
  return {res2.first, {res1.second, res2.second}};
}
void gd(int cn, int p, int tp) {
  for (auto nn : e[cn])
    if (nn != p) {
      d[nn][tp] = d[cn][tp] + 1;
      gd(nn, cn, tp);
    }
}
bool chk(int cn, int p, int lay) {
  if (lay == k) {
    if (e[cn].size() == 1) {
      return true;
    } else {
      return false;
    }
  }
  if (lay == 0) {
    if (e[cn].size() < 3) return false;
  } else {
    if (e[cn].size() < 4) return false;
  }
  for (auto nn : e[cn])
    if (nn != p) {
      if (!chk(nn, cn, lay + 1)) return false;
    }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int _ = 0; _ < (signed)(n - 1); _++) {
    int u, v;
    cin >> u >> v;
    e[u].push_back(v);
    e[v].push_back(u);
  }
  auto ree = diam();
  if (ree.first != k * 2) {
    cout << "No\n";
    return 0;
  }
  gd(ree.second.first, 0, 0);
  gd(ree.second.second, 0, 1);
  int cen = -1;
  for (int i = (1); i <= (signed)(n); i++) {
    if (d[i][0] == k && d[i][1] == k) {
      cen = i;
      break;
    }
  }
  if (chk(cen, 0, 0)) {
    cout << "Yes\n";
  } else {
    cout << "No\n";
  }
  return 0;
}
