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, x, y, pos, d[100000 + 1], r, i;
vector<int> v[100000 + 1];
pair<int, int> p;
queue<pair<int, int> > q;
void solve() {
  bool ok = false;
  while (!q.empty()) {
    p = q.front();
    q.pop();
    if (p.second == 1) {
      ok = true;
      break;
    }
    if (v[p.first].size() <= 3) break;
    for (i = 0; i < v[p.first].size(); i++) {
      if (d[v[p.first][i]] != 1) {
        q.push(make_pair(v[p.first][i], p.second - 1));
        d[v[p.first][i]] = 1;
      }
    }
  }
  if (ok == false || v[p.first].size() < 3) {
    cout << "NO" << endl;
    return;
  }
  while (!q.empty()) q.pop();
  for (i = 1; i <= n; i++) d[i] = 0;
  d[p.first] = 1;
  q.push(make_pair(p.first, 1));
  while (!q.empty()) {
    p = q.front();
    q.pop();
    if (p.second > k + 1) {
      cout << "NO" << endl;
      return;
    }
    if (v[p.first].size() == 1 && p.second != k + 1) {
      cout << "NO" << endl;
      return;
    }
    if (1 < p.second && p.second < k + 1) {
      if (v[p.first].size() <= 3) {
        cout << "NO" << endl;
        return;
      }
    }
    for (int i = 0; i < v[p.first].size(); i++) {
      if (d[v[p.first][i]] == 1) continue;
      d[v[p.first][i]] = 1;
      q.push(make_pair(v[p.first][i], p.second + 1));
    }
  }
  cout << "YES" << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (i = 1; i < n; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (i = 1; i <= n; i++) {
    if (v[i].size() == 1) {
      d[i] = 1;
      if (d[v[i][0]] != 1) {
        q.push(make_pair(v[i][0], k));
        d[v[i][0]] = 1;
      }
    }
  }
  solve();
}
