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;
const int dydis = 100001;
deque<int> masyvas[dydis];
deque<int> path;
deque<int> lpath;
int n, k;
int centras;
void longest(int atejau, int v, int atstumas) {
  path.push_back(v);
  if (path.size() > lpath.size()) lpath = path;
  for (int i = 0; i < masyvas[v].size(); i++) {
    if (masyvas[v][i] != atejau) longest(v, masyvas[v][i], atstumas + 1);
  }
  path.pop_back();
}
pair<int, int> virs;
void tolimiausias(int v, int atejau, int kel) {
  virs = max(virs, {kel, v});
  for (int i = 0; i < masyvas[v].size(); i++) {
    if (masyvas[v][i] != atejau) {
      tolimiausias(masyvas[v][i], v, kel + 1);
    }
  }
}
bool check(int v, int atejau, int level) {
  if (level > k) return false;
  if (level == k) {
    if (masyvas[v].size() == 1)
      return true;
    else
      return false;
  }
  if (v == centras) {
    if (masyvas[v].size() < 3) return false;
  } else {
    if (masyvas[v].size() < 4) return false;
  }
  bool grazinam = true;
  for (int i = 0; i < masyvas[v].size(); i++) {
    if (masyvas[v][i] != atejau) {
      grazinam = grazinam && check(masyvas[v][i], v, level + 1);
    }
  }
  return grazinam;
}
int main(int argc, const char* argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  if (n == 1 || k >= n) {
    cout << "No";
    return 0;
  }
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    masyvas[a].push_back(b);
    masyvas[b].push_back(a);
  }
  for (int i = 0; i < n; i++) {
    if (masyvas[i].size() == 1) {
      longest(-1, i, 0);
      break;
    }
  }
  centras = lpath[lpath.size() / 2];
  if (check(centras, -1, 0)) {
    cout << "Yes";
  } else
    cout << "No";
  return 0;
}
